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COMPUTATION  AND  MENSURATION 


THE  MACMILLAN  COMPANY 

NEW  YORK   •    BOSTON   •    CHICAGO 
ATLANTA  •    SAN  FRANCISCO 

MACMILLAN  &   CO.,  Limited 

LONDON   •    BOMBAY  •    CALCUTTA 
MELBOURNE 

THE  MACMILLAN  CO.  OF  CANADA,  Ltd. 

TORONTO 


COMPUTATION 
AND  MENSURATION 


BY 
P.    A.    LAMBERT,  M.A. 

PROFESSOR   OF   MATHEMATICS,    LEHIGH    UNIVERSITY 


NciM  fork 

THE   MACMILLAN   COMPANY 

1907 

AU  rights  reserved 


% 


Copyright,  1907, 
By  the  MACMILLAN  COMPANY. 

Set  up  and  electrotyped.    Published  September,  1907. 


J.  S.  Cushing  Co.  —  Berwick  &  Smith  Co. 
Norwood,  Mass.,  U.S.A. 


PREFACE 

The  transition  from  secondary  school  to  college  is 
disastrous  for  many  students.  This  is  due  largely  to  the 
fact  that  the  student  has  not  been  taught  to  make  inde- 
pendent use  of  what  he  has  learned.  The  transition 
should  be  accompanied  by  training  in  the  application  of 
the  knowledge  gained  in  the  secondary  school.  Such 
training  serves  a  threefold  purpose  —  it  makes  the 
student  realize  that  he  has  acquired  increased  power,  it 
reviews  the  work  of  the  secondary  school  in  an  interest- 
ing manner,  and  it  gives  an  outlook  into  the  work  of  the 
college. 

Such  a  transition  course  in  mathematics  the  author 
believes  may  be  based  on  this  text  on  Computation  and 
Mensuration.  This  transition  course  would  naturally 
come  at  the  end  of  the  secondary  school  course  or  at  the 
beginning  of  the  college  course.  * 

The  student  is  expected  to  refer  constantly  to  his  texts 
on  Algebra,  Geometry,  and  Trigonometry.  Formulas 
derived  in  all  elementary  text-books  are  neither  proved 
nor  tabulated  in  this  text.  The  aim  here  is  to  build  on 
the  foundation  already  laid. 


165869 


CONTENTS 

CHAPTER  I 
Approximate  Computation 

ARTICLE  PAGK 

1.  Approximate  Numbers 1 

2.  Direct  Measurement 3 

3.  Indirect  Measurement 3 

4.  Approximate  Multiplication 5 

5.  Approximate  Division 9 

CHAPTER  II 

Graphic  Computation 

6.  Graphic  Representation  of  Numbers 12 

7.  Addition  and  Subtraction  of  Segments 13 

8.  Multiplication  and  Division  of  Seginents       ....  14 
J^.  Square  Hoot  of  a  Segment 16 

10.  Graphic  Computation  of  Areas 17 

11.  Graphic  Solution  of  Triangles 18 

•  CHAPTER  III 
The  Method  of  Coordinates 

12.  The  Plane  Coordinates  of  a  Point 20 

13.  Area  of  a  Triangle  in  Coordinates  of  Vertices       ...      21 

14.  Areas  of  Any  Rectilinear  Figures 25 

CHAPTER  IV 
Volumes  of  Solids  bounded  by  Planes 

15.  The  Prismatoid 27 

16.  Special  Cases  of  Prismatoids  .        .        .        .        .      ,  .        .29 


Vlll  CONTENTS 


CHAPTER  V 


Computation  and  Use  of  Trigonometric  Functions 

ARTICLE  PAGE 

,17.  On  Angles 32 

18.  Trigonometric  Functions 34 

19.  Computation  of  Trigonometric  Functions      .         .        .         .36 

20.  On  Vectors 39 


CHAPTER  VI 
Computation  and  Use  of  Logarithms 


21.  Nature  of  Logarithms     . 

22.  Computation  of  Common  Logarithms 

23.  Arrangement  of  Tables  of  Logarithms 

24.  Computation  by  Means  of  Logarithms 

25.  The  Compound  Interest  Formula  . 

26.  The  Slide  Rule        .... 


44 
45 
50 
54 
59 
62 


CHAPTER  VII 

On  Limits  ^ 

27.  The  Infinite  Decreasing  Geometric  Progression    ...  65 

28.  The  Length  of  a  Curved  Line 66 

29.  The  Computation  of  tt 67 

30.  An  Important  Limit 69 

31.  Length  and  Area  of  Involute  of  Circle 70 

CHAPTER  VIII 

Graphic  Algebra 

32.  The  Graph  of  an  Equation 73 

33.  Equations  of  Lines 75 

34.  Graphic  Solution  of  Equations 76 

35.  Inequalities  treated  Graphically     .        •        .        .        .        .78 


CONTENTS  IX 


CHAPTER  IX 
Areas  bounded  by  Curves 

ARTICLE  PAGE 

36.  Exact  Areas    .         .        .         .  " 80 

37.  Approximate  Equation  of  a  Curve 82 

38.  Approximate  Areas 83 

CHAPTER  X 
Volumes  of  Solids 

39.  Exact  Volumes 87 

40.  Approximate  Volumes 88 

41.  Applicability  of  Prismoidal  Formula 90 


COMPUTATION  AND  MENSURATION 

CHAPTER  I 

APPROXIMATE  COMPUTATION 
Art.  1.  —  Approximate  Numbers 

Frequently  numbers  occurring  in  computation  are 
known  only  approximately. 

If  the  length  of  a  line  is  to  be  measured,  it  may  be 
absurd,  either  from  the  nature  of  the  problem  to  be  solved 
or  the  means  of  measurement  employed,  to  attempt  to  de- 
termine the  length  of  the  line  beyond  hundredths  of  a  foot. 

The  square  root  of  a  number  not  a  perfect  square  can- 
not be  expressed  exactly  in  numbers,  but  its  approximate 
value  to  any  desired  number  of  decimal  places  may  be  found. 

For  example,  the  approximate  values  of  V5  are  as  follows; 

to  one      decimal  place   2.2''" 
to  two     decimal  places  2.24" 
to  three  decimal  places  2.236"'' 
to  four    decimal  places  2.2361" 
to  five     decimal  places  2.23607" 
to  six       decimal  places  2.236068" 
to  seven  decimal  places  2. 2360680"  * 

*In  these  approximate  values  of  Vo  the  figure  in  the  last  decimal 
place  is  always  the  nearest  integer. 

B  1 


2  COMPUTATION    AND    MENSURATION 

In  each  approximate  value  the  sign  +  or  —  written 
after  the  last  figure  indicates  that  the  error  must  be 
respectively  added  or  subtracted.  The  error  in  every 
case  is  not  greater  than  5  units  of  the  decimal  place  next 
following  the  last  decimal  place  of  the  approximation. 

The  approximate  value  of  V5  to  eight  decimal  places 
cannot  be  determined  until  the  figure  in  the  ninth  decimal 
place  has  been  computed. 

The  ratio  of  the  circumference  of  a  circle  to  its  diam- 
eter cannot  be  expressed  exactly  in  numbers,  but  its  value 
has  been  computed  to  707  decimal  places.  This  ratio  is 
denoted  by  tt,  and  its  value  to  15  decimal  places  is 

7r=  3.141592653589793. 

The  value  of  tt  most  commonly  used  is  3.1416. 

Problem  l.   Find  the  limit  of  error  when  ^^-  is  used 

for  TT. 

Pkoblem  2.    Find  the  limit  of  error  when  ^^J  is  used 

for  TT. 

Problem  3.  Find  the  limits  of  error  of  the  sums  of 
V26,  V27,  V28,  V29,  each  computed  approximately  to 
four  decimal  places*. 

Problem  4.  The  yard  is  defined  by  law  as  g^^  meter.* 
Find  the  number  of  meters  in  a  yard  correct  to  4  decimals. 

Problem  5.  The  pound  is  defined  by  law  as  2:204^2  kilo- 
gram.* Find  the  number  of  kilograms  in  a  pound  correct 
to  4  decimals. 

*  By  decision  oi  United  States  Superintendent  of  Weights  and  Meas- 
ures,^ on  April  5,  1903,  with  the  approval  of  the  Secretary  of  the  Treasury. 


APPROXIMATE    COMPUTATION  3 

Art.  2.  —  Direct  Measurement 

The  direct  measurement  of  the  length  of  a  line  is 
effected  by  actually  applying  to  the  line  the  standard  unit 
of  length.  However  carefully  several  direct  measure- 
ments are  made,  the  results  are  generally  not  the  same. 

By  universal  agreement,  the  arithmetic  mean  of  the  re- 
sults of  several  direct  measurements,  made  with  the  same 
care,  is  accepted  as  the  best  result  to  be  obtained  from 
these  measurements. 

Problem  6.  Eight  measurements  of  a  line  give  the 
values  186.4,  186.3,  186.2,  186.3,  186.2,  185.9,  and  186.4 
inches.  Find  the  best  value  of  the  length  of  the  line 
based  on  these  measurements. 

Problem  7.  Four  measurements  of  an  angle  give  62° 
25'  10". 0  ;  62°  25'  1".5  ;  62°  25'  23". 3,  and  62°  25'  30". 0. 
Find  the  best  result  determined  by  these  measurements. 

Problem  8.  Four  measurements  of  a  base  line  give 
1472.34,  1471.99,  1472.25,  and  1472.14  feet.  Find  the 
best  approximate  length  of  the  line  to  be  obtained  from 
these  measurements. 

Art.  3.  —  Indirect  Measurement 

Example  l.  Suppose  A,  B,  C  to  be  three  stations  form- 
ing the  vertices  of  a  right  triangle,  and  let  it  be  required 
to  determine  the  length  of  the  hypotenuse  AB  when  an 
obstruction  between  A  and  B  makes  the  direct  measure- 
ment of  this  length  impracticable. 

If    by    direct    measurement    ^C=  1275.43    feet    and 


4  COMPUTATION  And  mensuration 

BC=  1526.56  feet,  AB  may  be  computed  by  means  of  the 
proposition  in  Geometry  — 

The  square  on  the  hypotenuse  equals  the  sum  of  the 
squares  on  the  other  two  sides,  or  expressed  as  a  formula, 

AB'^  =  AO^  +  BlP. 

The  number  expressing  the  length  of  AB  is  found  by 
extracting  a  square  root  and  may  be  carried  out  to  any 
desired  number  of  decimal  places;  but  since  the  measured 
lines  AC  and  BC  are  determined  only  to  two  decimals,  it 
would  be  absurd  to  attempt  to  determine  by  computation 
the  length  of  AB  beyond  two  decimals. 

Example  2.  Let  it  be  required  to  determine  the  cir- 
cumference of  a  circle  whose  diameter  is  found  by  direct 
measurement  to  be  425.63  feet. 

The  circumference  of  this  circle  is  tt  x  425.63  feet. 
Since  the  diameter  of  the  circle  is  given  to  two  decimals, 
it  is  impossible  to  determine  by  computation  the  circum- 
ference beyond  two  decimal  places.  It  is  therefore  absurd 
to  use  more  decimal  places  in  the  value  of  tt  than  are  neces- 
sary to  determine  the  product  tt  x  425.63  to  two  decimal 
places.  What  is  needed  is  a  method  of  approximate  com- 
putation which  will  determine  the  circumference  correct  to 
two  decimal  places  with  the  least  amount  of  computation. 

Mensuration  is  chiefly  concerned  with  problems  of 
indirect  measurement. 

Problem  9.  In  a  right  triangle  the  sides  about  the 
right  angle  measure  875.27  feet  and  56b A5  feet.  Com- 
pute the  length  of  the  hypotenuse. 


APPROXIMATE    COMPUTATION  5 

Problem  lO.  The  diameter  of  a  circle  measures 
365.18  feet.     Compute  the  length  of  the  circumference. 

Problem  ll.  The  circumference  of  a  circle  measures 
1000  feet.     Compute  the  diameter. 

Art.  4.  —  Approximate  Multiplication 

Example.  Let  it  be  required  to  compute  the  product 
TT  ><  425.63  correct  to  two  decimal  places. 

The  computation  is  carried  out  to  three  decimal  places 
in  order  that  the  second  decimal  place  may  be  corrected. 

The  first  step  is  to  determine  how  many  decimal  places 
in  the  value  of  ir  must  be  used  to  give  three  decimal 
places  in  the  product. 

The  computation  consists  in  multiplying  the  approx- 
imate value  of  TT  by  each  figure  of  425.63  and  adding  the 
partial  products.  These  partial  products  are  to  contain 
three  decimal  places  each.    • 

In  order  that  the  partial  product  of  the  approximate 
value  of  TT  by  the  units'  figure  of  425.63  shall  contain  three 
decimal  places,  the  value  of  tt  must  be  written  to  three 
decimal  places. 

For  each  integral  place  to  the  left  of  units'  place  in 
425.63  the  value  of  ir  must  be  extended  one  decimal  place 
to  the  right  in  order  that  the  corresponding  partial 
product  shall  contain  three  decimal  places. 

In  this  example,  therefore,  the  value  of  ir  must  be 
written  to  five  decimal  places,  and  in  order  that  the  partial 
product  by  the  figure  in  the  hundredths'  place  in  425.63 
may  be  corrected  in  the  third  decimal  place  the  value  of 
TT  is  written  to  six  decimal  places. 


6  COMPUTATION    AND    MENSURATION 

The  dot  placed  over  the  figure  in  the  fifth  decimal  place 
of  TT  indicates  that  this  figure  is  the  last  figure  to  the  right 
in  the  value  of  tt  to  be  used  to  form  the  partial  product 
corresponding  to  the  first  figure  to  the  left  of  425.63. 

As  the  figure  of  the  multiplier  for  which  a  partial  prod- 
uct is  to  be  computed  moves  one  place  to  the  right,  the 
last  figure  of  the  multiplicand  to  be  used  moves  one  place 
to  the  left. 

There  is  considerable  advantage  in  forming  the  partial 
products  for  the  figures  of  425.63  from  right  to  left.  The 
last  figures  of  the  partial  products  are  corrected  for  the 
rejected  part  of  the  value  of  tt.  The  signs  +  and  —  in- 
dicate the  •  direction  of  the  errors  committed  by  these 
corrections.  The  sign  +  indicates  that  the  error  is  to  be 
added,  the  sign  —  that  the  error  is  to  be  subtracted. 

The  numerical  value  of  the  error  cannot  exceed  J  of  a 
unit  of  the  third  decimal  place. 

The  computation  now  appears  as  follows : 

3.141593 
425.63 


1256.637  + 

62.832- 

15.708- 

1.885- 

.094+ 

1337.156 


The  error  of  this  result  must  lie  between  —  IJ  and  + 1 
units  of  the  third  decimal  place.  It  is  therefore  uncertain 
if  the  figure  in  the  second  decimal  place  is  5  or  6. 


APPROXIMATE    COMPUTATION  7 

It  is  necessary  in  this  particular  problem  to  carry  out 
the  approximate  multiplication  to  4  decimals  as  follows : 

3.1415927 

425.63 

1256.6371- 

62.8319- 

15.7080- 

1.8850- 

.0942  + 


1337.1562 

The  error  of  this  result  must  lie  between  —  2  and  +  J 
units  of  the  fourth  decimal  place.  Hence  the  product 
correct  to  two  places  of  decimals  is  1337.16. 

The  following  arrangement  of  this  computation  is  per- 
haps to  be  preferred. 

Place  the  units'  figure  of  425.63  under  the  third  decimal 
place  of  the  value  of  tt  and  then  write  in  inverse  order 
the  figures  of  425.53  under  the  figures  of  the  approximate 
value  of  TT.  This  brings  each  figure  of  the  multiplier  in 
line  with  that  figure  of  the  multiplicand  which  is  used  to 
begin  the  corresponding  partial  product.  The  above  ap- 
proximate multiplications  appear  as  follows  : 

3.141593  3.1415927 

36.524  36.524 


1256  637-^ 

12566  371- 

62  832- 

628  319- 

15  708- 

157  080- 

1885- 

18  850- 

94+ 

942  + 

1337.156 

1337.1562 

8  COMPUTATION    ANB    MENSURATION 

Problem  12.  Find  the  circumference  of  a  shaft  whose 
radius  is  4.32  inches. 

Pkoblem  13.  If  a  yard  is  1.904  meters,  find  the  num- 
ber of  meters  in  23.463  yards. 

Problem  14.  Find  the  interest  on  11525.75  for  one 
year  at  6|%,  correct  to  f  0.01. 

Problem  15.  The  distance  of  the  moon  from  the  earth 
is  59.97  times  the  earth's  radius.  If  the  earth's  radius 
is  3962.824  miles,  find  the  distance  to  the  moon  correct 
within  1  mile. 

Problem  16.  Find  the  area  of  the  circle  whose  radius 
is  16:27  feet. 

Problem  17.  Find  the  volume  of  the  sphere  whose 
radius  is  3.53  feet. 

Problem  18.  Find  the  volume  of  the  cone  whose  base 
is  a  circle,  radius  2.35  feet,  altitude  5.75  feet. 

Problem  19.  Find  the  volume  of  the  rectangular  par- 
allelopiped  whose  dimensions  are  8.53  feet,  6.27  feet,  and 
4.65  feet. 

Problem  20.  Find  the  length  of  the  diagonal  of  the 
parallelopiped  in  Problem  19. 

Problem  21.    Compute  168  V3  to  two  decimal  places. 

Problem  22.    Compute  tt  \/2  to  three  decimal  places. 

Problem  23.    Compute  tt^  to  four  decimal  places. 


APPROXIMATE    COMPUTATION  9 

Art.  5. — Approximate  Division 

Example.  Let  it  be  required  to  compute  correct  to 
two  places  of  decimals  the  diameter  of  the  circle  whose 
circumference  measures  587.35  feet. 

In  order  that  the  figure  in  the  second  decimal  place  of 
the  quotient  may  be  controlled,  the  quotient  of  587.35 
divided  by  tt  is  computed  to  three  places  of  decimals. 
By  inspection  the  quotient  has  three  integral  places. 
Hence  six  figures  of  the  quotient  are  to  be  computed.  The 
nature  of  the  problem  shows  that  this  computation  requires 
six  figures  of  the  divisor  and  six  figures  in  the  dividend. 

The  successive  partial  products  are  formed  as  in  the 
process  of  approximate  multiplication  in  the  preceding 
article.     The  computation  is  arranged  as  follows : 


587.350 

3.14159+ 

314  159  + 

186.959 

273  191 

251  327  + 

21864 

18  850- 

3  014 

2  827- 

187 

157  + 

30 

28  + 
2 

The  signs  +  and  —  placed  after  the  partial  products 
indicate  the  direction  of  the  error  of  the  partial  products 


UNIVERSITY 


10  COMPUTATION    AND    MENSURATION 

The  error  in  the  sum  of  the  partial  products  must  be 
between  -j-  2  and  —  1  units  of  the  last  decimal  place  of 
the  dividend.  Hence  the  last  remainder  lies  between  0 
and  -h  3  of  these  units,  and  the  error  of  the  quotient  lies 
between  0  and  +  1  units  of  the  third  decimal  place. 
Hence  the  quotient  correct  to  two  decimal  places  is  186.96. 
There  is  some  advantage  in  arranging  the  figures  of  the 
quotient  in  such  a  manner  that  the  partial  products  begin 
with  the  product  of  the  figure  of  the  quotient  and  the 
figure  of  the  divisor  directly  over  it.  The  computation 
then  takes  this  form, 


587.350 

3.14159  + 

314  159+ 

959.681 

273  191 

251  327  + 

21  864 

18  850- 

3  014 

2  .827- 

187 

157  + 

30 

28  + 
2 

Problem  24.  Find  the  radius  of  the  circle  whose  cir- 
cumference is  425.76  feet. 

Problem  25.  Find  the  area  of  the  circle  whose  circum- 
ference is  628.32  feet. 

Problem  26.  Find  the  circumference  of  the  circle 
whose  area  is  3848.45  square  feet. 


APPROXIMATE    COMPUTATION  11 

Problem  27.  Find  the  diameter  of  a  wheel  which 
makes  373  revolutions  in  a  mile. 

Problem  28.  How  many  revolutions  per  mile  are 
made  by  a  locomotive  drive  wheel  4.5  feet  in  diameter  ? 

Problem  29.    Compute  — -  to  four  decimal  places. 

Problem  30.  Find  correct  to  four  decimal  places  the 
number  of  degrees  in  an  angle  of  1  radian. 

Problem  31.  The  moon  revolves  about  the  earth  in  28 
days,  7  hours,  43  minutes,  11.5  seconds.  What  is  the 
average  angle  passed  over  in  a  day  ? 


CHAPTER  II 

GKAPHIO  COMPUTATION 
Art.  6.  —  Graphic  Eepresextatiox  of  Numbers 

On  a  straiglit  line  mark  a  series  of  equidistant  points. 
Select  one  of  these  points  and  call  it  zero.  Starting  from 
zero  point  call  the  successive  points  in  one  direction  +  1, 
+  2,  +  3,  +  4,  4-  5,  •••  ,  the  successive  points  in  the  oppo- 
site direction  —  1,  —  2,  —  3,  —  4,  —  5,  •••  .  There  is  thus 
attached  to  each  integral  number,  positive  or  negative,  one 
point  of  the  line. 

Insert  between  the  points  attached  to  two  consecutive 
integral  numbers,  3  and  4,  for  example,  nine  equidistant 
points  and  call  these  points  3.1,  3.2,  3.3,  3.4,-8.5,  3.6,  3.7, 
3.8,  3.9.  There  is  thus  attached  to  each  number  contain- 
ing not  more  than  one  decimal  place  one  point  of  the  line. 

-8-7-6-5-4-3-2-1       9       1      ^      3      4      5       6       7      8 
H i 1 1 — i — t — I — I 1 1 1 1 1 — i 1 1 H 

,  Fig.  1. 

Theoretically,  this  operation  of  inserting  between  the 
points  attached  to  two  consecutive  numbers  nine  equidis- 
tant points  may  be  repeated  indefinitely,  but  practically 
little  is  gained  by  attempting  to  attach  points  to  numbers 
containing  more  than  one  decimal  place. 

Conversely,  if  any  point  is  taken  at  random  on  the 
straight  line,  the  number  attached  to  the  point  can  be  de- 

12 


GUAPHIC    COMPUTATION  13 

teriiiined  to  one  decimal  place  with  considerable  certainty, 
but  the  attempt  to  determine  the  number  beyond  one  dec- 
imal place  would  be  accompanied  by  great  uncertainty. 

If  the  line  segment  joining  the  points  attached  to  two 
consecutive  whole  numbers  is  taken  as  the  unit  of  length, 
the  number  attached  to  any  point  of  the  line  measures  the 
length  of  the  line  segment  from  the  zero  point  to  this  point. 
The  algebraic  sign  of  the  number  indicates  the  direction 
in  which  the  line  segment  extends  from  the  origin. 

The  straight  line  with  numbers  attached  to  successive 
equidistant  points  now  becomes  a  scale  for  measuring  the 
lengths  of  line  segments. 

Art.  7. — Addition  and  Subtraction  of  Segments 

A  line  segment  is  a  definite  part  of  a  straight  line  and 
extending  in  a  definite  direction  along  the  line. 

A  line  segment  has  an  initial  point  and  a  terminal  point. 
The  segment  extends  from  the  initial  point  to  the  terminal 
point. 

A  line  segment  is  transferred  from  one  straight  line  to 
another  by  means  of  dividers. 

Two  line  segments  are  equal  when  one  can  be  trans- 
ferred into  the  other. 

When  line  segments  are  to  be  added  or  subtracted  they 
are  transferred  to  the  same  straight  line. 

To  add  two  line  segments  having  the  same  sign,  lay  off 
the  two  segments  on  a  straight  line  both  in  the  same 
direction,  making  the  terminal  point  of  the  first  segment 
the  initial  point  of  the  second.     The  line  segment  which 


14  COMPUTATION   AND    MENSURATION 

now  extends  from  the  initial  point  of  the  first  segment 
to  the  terminal  point  of  the  second  is  the  sum  of  the 
two. 

To  add  two  line  segments  having  opposite  signs,  lay  off 
the  two  segments  on  a  straight  line  in  opposite  directions, 
making  the  terminal  point  of  the  first  segment  the  initial 
point  of  the  second.  The  line  segment  which  now  extends 
from  the  initial  point  of  the  first  segment  to  the  terminal 
point  of  the  second  is  their  sum. 

To  subtract  two  line  segments,  change  the  sign  of  the 
segment  to  be  subtracted  and  add  to  the  other  segment. 

To  add  or  subtract  numbers  graphically,  find  from  the 
scale  of  segments  the  line  segments  corresponding  to  the 
given  numbers,  then  find  the  sum  or  difference  of  these 
line  segments  and  from  the  scale  determine  the  corre- 
sponding number. 

Art.  8,  —  Multiplication  and  Division  of  Segments 

To  find  the  product  of  two  line  segments,  define  the 
product  by  the  proportion 

product  :  multiplicand  =  multiplier  :  unity. 

The  product  of  two  line  segments  is  therefore  the  line 
segment  which  is  the  fourth  proportional  to  the  two  given 
line  segments  and  the  unit  line  segment. 

This  fourth  proportional  is  constructed  by  drawing' two 
straight  lines  through  a  common  point  making  any  con- 
venient angle,  the  positive  direction  on  each  line  being 
indicated  by  the  arrowhead,  and  laying  oft'  from  the 
common  point  the  unit  segment  and  multiplier  on  the  first 


GRAPHIC    COMPUTATION  15 

straight  line,  the  multiplicand  on  the  second  straight  line. 
Now  join  the  terminal  points  of  the  unit  segment  and 
the  multiplicand  by  a  straight  line  and  draw  through  the 
terminal  point  of  the  multiplier  a  parallel  to  this  straight 


line.  The  point  of  intersection  of  this  parallel  with  the 
second  straight  line  is  the  terminal  point,  the  common  point 
the  initial  point  of  the  product  segment.     In  the  figure 

1^=^,  hence  00=  OA  .  OB. 

To  find  the  quotient  of  two  line  segments,  define  the 
quotient  by  the  proportion 

quotient  :  1  =  dividend  :  divisor. 

The  quotient  of  two  line  segments  is,  therefore,  the  line 
segment  which  is  the  fourth  proportional  to  the  unit  line 
segment  and  the  two  segments  which  are  respectively  the 
dividend  and  the  divisor. 

This  fourth  proportional  is  constructed  by  drawing  two 
straight  lines  through  a  common  point,  making  any  con- 
venient angle,  laying  off  from  the  common  point  the 
dividend  and  divisor  on  the  first  straight  line,  the  unit 
segment  on  the  second  straight  line.  Now  join  the  ter- 
minal points  of  the  unit  segment  and  the  divisor  by  a 
straight  line  and  draw  through  the  terminal  point  of  the 


16  COMPUTATION    AND    MENSURATION 

dividend  a  parallel  to  this  straight  line.     The  point  where 
this  parallel  intersects  the  second  straight  line  is  the  ter- 


minal point,  the  common  point   the  initial  point  of  the 

quotient  segment. 

In  the  figure 

00      OA  ,  __  ^^      OA 
—  =  —,  hence  0(7=—. 

To  find  graphically  the  product  or  the  quotient  of  two 
numbers,  find  by  means  of  the  scale  of  segments  the  line 
segments  corresponding  to  the  given  numbers. 

Then  find  the  product  or  quotient  of  these  line  seg- 
ments and  from  the  scale  the  number  corresponding  to  the 
product  or  quotient  segment. 

Problem  32.  Find  graphically  the  product  of  8.6  and 
7.3. 

Problem  33.  Find  graphically  the  quotient  of  47.5 
divided  by  11.7. 

Art.  9.  —  Square  Root  of  a  Segment 

If  the  product  of  two  equal  line  segments  equals  the 
given  line  segment,  one  of  the  equal  segments  is  called 
the  square  root  of  the  given  line  segment. 


GRAPHIC    COMPUTATION  17 

To  construct  the  line  segment  which  is  the  square  root 
of  the  given  line  segment,  find  in  any  manner  two  line 
segments  whose  product  equals  the  given  line  segment. 
On  the  sum  of  these  two  line  segments  as  a  diameter,  con- 
struct a  semicircumference.  The  perpendicular  to  the 
diameter  at  the  point  of  meeting  of  the  two  line  segments 


Fig.  4. 

and  terminating  in  the  circumference  is  the  line  segment 
required. 

Repeated  applications  of  this  operation  make  it  possible 
to  extract  a  root  which  is  any  power  of  2  by  the  use  of 
the  straight  line  and  circle. 

Problem  34.    Find  graphically  V24. 

Problem  35.    Add  graphically  Vl9,  V2i,  and  V28. 

Art.  10.  —  Graphic  Computation  of  Areas 

Construct  as  accurately  as  practicable  the  figure  whose 
area  is  to  be  found  on  cross-section  paper,  taking  the  dis- 
tance between  consecutive  lines  to  represent  the  unit  of 
length  in  the  measurement  of  the  figure.  The  square  of 
the  cross-section  paper  now  represents  the  unit  of  area, 
and  the  area  of  the  figure  is  determined  by  counting  the 
squares  wholly  within  the  boundary  line  of  the  figure  and 


18  COMPUTATION    AND    MENSURATION 

estimating  the  squares  partially  within  the  boundary  line. 
The  accuracy  of  the  result  depends  upon  the  skill  in 
estimating  the  sum  of  the  partial  squares. 

Problem  36.    Find  graphically  the  area  of  the  triangle 
whose  sides  are  8,  12,  15  feet. 

Problem  37.    Find  graphically  the  area  of  the  circle 
whose  diameter  is  9  inches. 

Problem  38.    Find  graphically  the  area  of  the  equi- 
lateral triangle  whose  side  is  7  feet. 


Art.  11.  —  Graphic  Solution  of  Triangles 

The  data  necessary  for  the  solution  of  a  triangle  is 
sufficient  for  the  construction  of  the  triangle. 

If  the  triangle  is  constructed  with  the  given  parts  as 
accurately  as  practicable  by  the  aid  of  straight  edge,  pro- 
tractor, and  dividers,  by  means  of  these  same  instruments 
the  values  of  the  unknown  parts  of  the  triangle  can  be 
found  with  considerable  accuracy. 

Problem  39.  Two  sides  of  a  triangle  are  7.8  feet  and 
12.5  feet,  their  included  angle  is  28°  W,  Find  the  re- 
maining three  parts  graphically. 

Problem  40.  Two  sides  of  a  triangle  are  9.5  feet  and 
16.3  feet.  The  angle  opposite  the  first  side  is  16°  15^ 
Find  the  other  parts  graphically. 

Problem  41.  One  side  of  a  triangle  is  10.6  feet,  the 
adjacent  angles  are  Qb°  lb'  and  80°  10^  Find  graphically 
the  other  two  sides. 


GRAPHIC    COMPUTATION  19 

Peoblem  42.  The  sides  of  a  triangle  are  18  feet,  25 
feet,  and  29  feet.     Find  the  angles  graphically. 

Problem  43.  Two  sides  of  a  triangle  are  4.6  feet  and 
12.8  feet.  The  angle  opposite  the  former  sides  is  60°  30'. 
Find  the  remaining  parts  graphically. 


CHAPTER   III 


THE  METHOD  OP  OOORLINATES 

Art.  12. — The  Plane  Coordixates  of  a  Poixt 

In  a  given  plane  draw  two  straight  lines  perpendicular 
to  each  other.  On  each  of  these  lines,  with  their  inter- 
section at  zero  point,  construct  a  linear  scale,  denoting  the 
Y  positive     direction 


(-3,2) 


(3.2) 


on  each  line  by  an 
arrowhead.        Call 
these  two  lines   of 
reference    the    X- 
-^  X    axis  and  the  y-axis. 
Select  any  point 
P  in  the  plane,  and 
through  this  point 
draw  a  line  paral- 
FiG.  5.  lei  to  the  I^-axis  to 

meet  the  X-axis.  The  number  which  expresses  the  distance 
and  direction  of  this  meeting  point  from  the  zero  point 
is  called  the  abscissa  of  the  point  P  and  is  denoted  by  x. 
Through  the  point  P  draw  a  parallel  to  the  X-axis  to 
meet  the  F-axis.  The  number  which  expresses,  the  dis- 
tance and  direction  of  this  meeting  point  from  the  zero 
point  is  called  the  ordinate  of  the  point  and  is  denoted  by  y. 
The  abscissa  of  a  point  may  be  defined  as  the  number 
which  expresses  the  distance  and  direction  of  the  point 

20 


THE   METHOD    OF    COORDINATES  21 

from  the  y-axis  measured  on  a  parallel  to  the  X-axis  ; 
and  the  ordinate  may  be  defined  as  the  number  which  ex- 
presses the  distance  and  direction  of  the  point  from  the 
X-axis  measured  on  a  parallel  to  the  y-axis. 

The  abscissa  of  a  point  determines  a  straight  line  par- 
allel to  the  y-axis  in  which  the  point  must  be  situated. 

The  ordinate  of  a  point  determines  a  straight  line  par- 
allel to  the  X-axis  in  which  the  point  must  be  situated. 

The  abscissa  and  ordinate  of  a  point  together  determine 
the  point  in  the  plane  and  are  called  the  coordinates  of 
the  point. 

The  point  whose  coordinates  are  a:  =  3,  ?/  =  —  2  is  called 
the  point  (3,  -  2). 

Problem  44.  Where  is  the  point  situated  whose  ab- 
scissa is  4? 

Problem  45.  Where  is  the  point  situated  whose  ordi- 
nate is  —  5  ? 

Problem  46.    Locate  the  points  (3,  5),  (—4,  6),  (0,  8). 

Problem  47.  Construct  the  triangle  whose  vertices 
are  the  points  (10,  4),  (6,  -  8),  (15,  12). 

Problem  48.  Locate  the  points  (5,  8),  (10,  7)  and  find 
the  distance  between  them. 

Problem  49.  Find  the  area  of  the  triangle  whose  ver- 
tices are  (3,  7),  (8,  5),  (13,  15). 

Art.  13. — Area  of  Triangle  in  Coordinates  of  Vertices 
Suppose  the  vertices  of   the  triangle  to  be  the  points 
A  (^r  ^i)'  A  (^2'  ^2)'  ^3  C^3'  ^3)- 


22 


COMPUTATION    AND    MENSURATION 


Two  cases  will  be  considered,  Case  I,  when  all  the  coor- 
dinates of  the  vertices  are  positive,  Case  II,  when  some  of 
these  coordinates  are  negative. 

It  must  be  remembered  that  line  segments  parallel  to 
the  X-axis  are  positive  when  they  extend  in  the  positive 
direction  of  the  X-axis,  and  line  segments  parallel  to  the 
y-axis  are  positive  when  they  extend  in  the  positive 
direction  of  the  Z-axis. 

In  the  proposition 

The  area  of  a  trapezoid  is  half  the  sum  of  the  parallel 
bases  by  the  altitude, 

the  bases  and  altitude  are  positive  line  segments. 
Y 


Fig.  6. 


Case  I.  The  area  of  the  triangle  P^P^P^  equals  the  sum 
of  the  areas  of  the  trapezoids  P^D^P^P^^  and  P^B^D^P^ 
minus  the  area  of  the  trapezoid  P^P^P^P^. 

Hence 
2  Area  triangle  P^P^P^  =  B^P^  x  QB^P^  -f-  B^P^) 

-F  i>3Z>2  X  (DgPg  +  D2P2)  -  -^1^2  X  (i>iPi  +  AA)- 
Now  x^  =  OB^,  x^  =  OB^,  x^  =  ODg, 


Vi  =  ^i^v  Vi  =  ^2^T  Vz  =  A^a 


THE   METHOD    OF    COORBINATES 


23 


Hence  DiD^  =^s~~  ^v  -^s^2  =  ^2  ~  ^3^  -^1^2  =  ^2  ~  ^v  ^^^ 
2  Area  P^P^P^  =  (2-3  -  x^)  (y^  +  7/^) 

+  (^2  -  ^3)  C2/3  +^2)  -  (^2  -  ^1)  (^1  +  ^2)- 

Multiplying  out  and  simplifying, 

2  Area  P^P^P^  =  x^y^^  +  ^^^2^/3  +  ^3^1  -  ^xVz  -  ^2^/1  -  ^3^2- 

This  expression  for  double  the  area  of  a  triangle  in 
terms  of  the  coordinates  of  its  vertices  is  readily  repro- 
duced by  arranging  the  coordinates  of  the  vertices 
^1^1   in  order  in  two  vertical  columns  and  repeating  at 
^2^2   the  end  of  the  column  the  coordinates  of  the  first 
^^  point.     The  three  positive  terms  of  the  area  are  the 
^  ^  products  of  each  of  the  three  abscissas  by  the  next 
lower  ordinate,  and  the  three  negative  terms  of  the  area 
are  the  products  of  each  of   the  three  ordinates   by  the 
next  lower  abscissa. 

The  area  of  the  triangle  is  positive  when  the  perimeter 
is  supposed  to  be  described  in  such  a  manner  that  the  area 
lies  to  the  left  of  the  boundary  line. 


24  COMPUTATION   AND    MENSURATION 

Case  II.  The  area  of  the  triangle  P^P^P^  equals  the 
area  of  the  trapezoid  P^E^E^P^  minus  the  sum  of  the 
areas  of  the  triangles  P^E^P^  and  P^E^Py 

Hence 

2  Area  triangle  P.P^P,  =  (P^E,^  +  ^3^3)  x  A^a 

-  P^E^  X  E^P,  -  P^E,  X  P,E,, 

Now  x^  =  OZ>i,  x^=  —B^ 0,  x^  =  02>3, 

^1  =  A^r  ^2  =  -  A^2'  3/3  =  -  ^3^3- 

Hence  P^E^=  - y^ ^y^^  P^E^  =  -  «/3  +  Vv 

D^D^  =  -x^-h  x^,  E^P^  =  —x^  +  x^,  P^E^  =  ^3  -  ^v  and 
2  Area  P^P^P^  =  (2  ^1  -  ^2  -  2/3)  (  ^  ^1  +  ^3) 

-  (-  3/2  +  2/1)  (-  ^2  +  ^1)  -  (-  ^3  +  ^1)  (^3  -  ^i)- 

This  reduces  to 

2  Area  P^P^P^  =  ^1^2  +  ^2^/3  +  ^3^1  -  ^1^2  -  V'i^z  -  Vz^v 
the  same  result  as  was  found  for  Case  I. 

This  expression  for  double  the  area  of  a  triangle  in 
terms  of  the  coordinates  of  its  vertices  is  entirely  general. 

Problem  50.  Find  the  area  of  the  triangle  whose 
vertices  are  the  points  (12,  —  5),  (  —  8,  7),  (10,  15),  the 
coordinates  being  measured  in  inches. 

Problem  51.  Find  the  area  of  the  triangle  the  co- 
ordinates of  whose  vertices  measured  in  chains  are  (15.75, 
4.26),  (18.25,  20.63),  (21.43,  16.52). 

Problem  52.    Show  that  the  points  (1,  4),  (3,  2),  (—3, 

8)  lie  in  a  straight  line. 


THE   METHOD    OF    COORDINATES 


25 


Art.  14.  —  Areas  of  Any  Eectilinear  Figures 

Let  PiCx^^i).  ^(^2^2)'  ^3(^3^3)'  -^4(^4^4)  be  the 
vertices  of  a  quadrilateral.  The  straight  line  P1P3 
divides  this  quadrilateral  into  the  triangles  P^P^P^  and 

2  Area  P^P^P^  =  x^y^  +  2:2^3  +  x^^  -  y^x^^  -  y^x^  -  y^x^. 
2  Area  P^P^P^  =  x^y^  +  x^^  +  x^y^  -  y^x^  -  y^x^  -  y^Xy 

Hence 

2  Area  P^PJP^P^  =  x^y^-\-  x^^  +  x^^  +  x^y^  -  y^x^ 

-  ^2^3  -  ^3^4  -  ^4^1- 


^1^1 
^22/2 

^4^4 


This  expression  for  double  the  area  of  a  quadrilateral  is 
readily  reproduced  by  writing  the  coordinates  of  the  ver- 
tices in  regular  order  in  two  vertical  columns,  the  co- 
ordinates of  the  first  vertex  being  repeated  at  the  end  of 
the  cqlumns.;  The  positive  terms  of  the  expression  for 
the  double  area  are  the  products  of  each  abscissa  by  the 
next  lower  ordinate  ;  the  negative  terms  are  the  products 
of  each  ordinate  by  the  next  lower  abscissa. 


26 


COMPUTATION  And  mensuration 


This  same  method  gives  the  area  in  terms  of  the  co- 
ordinates of  its  vertices  of  any  rectilinear  figure. 

Peoblem  53.  The  coordinates  measured  in  chains  of 
the  angular  points  of  a  quadrilateral  field  are  (10.25,  0), 
(3.21,  7.35),  (  -  8.75,  0),  (2.37,  -  9.13).  Construct  the 
figure  and  find  the  area  of  the  field  in  acres. 

Pkoblem  54.  Compute  the  area  and  construct  the 
figure  of  the  polygon  the  coordinates  of  whose  angular 
points  measured  in  feet  are 

(  -  27,  0),  (20,  3.75),  (16.5,  4),  (15,  7.5),  {^.Q^ 

Problems  55,  56,  and  57.  The  coordinates  measured 
in  chains  of  the  successive  corners  of  three  tracts  of  land 
bounded  by  straight  lines  are  as  follows.  Compute  the 
areas  and  construct  the  figures. 


Problem  55 

Problem  56 

Problem  57 

X 

y 

X 

y 

X 

y 

0 

-8.12 

13.31 

6.30 

-30 

90 

9.31 

-  16.21 

24.09 

7.21 

0 

140 

25.11 

-9.61 

14.11 

12.40 

80 

130 

30. 

0 

26. 

13.10 

50 

90 

23. 

0 

26.23 

15. 

54 

80 

22.1 

15.4 

20. 

20.37 

100 

40 

7.23 

16.48 

10. 

16.20 

60 

20 

0 

11. 

1.01 

16. 

10 

0 

0 

14.30 

5 

70 

CHAPTER   IV 

VOLUMES   OP  SOLIDS  BOUNDED  BY  PLANES 
Art.  15.  —  The  Prismatoid 

The  polyhedron  two  of  whose  faces  are  any  two  poly- 
gons in  parallel  planes  and  whose  other  faces  are  composed 
of  triangles  formed  by  so  joining  the  vertices  of  these  par- 
allel polygons  that  each  line  in  order  forms  a  triangle 
with  the  preceding  line  and  one  side  of  either  parallel 
polygon  is  called  a  prismatoid. 

The  parallel  polygons  are  called  the  bases  of  the  pris- 
matoid. The  section  of  the  prismatoid  formed  by  a  plane 
midway  between  the  bases  is  called  the  midsection.  The 
perpendicular  distance  between  the  bases  is  called  the 
altitude  of  the  prismatoid. 

Call  the  altitude  of  the  prismatoid  A,  the  areas  of  the 
bases  h^  and  ^2^  the  area  of  the  midsection  m.  Let  0  be 
any  point  in  the  midsection.  Draw  lines  from  0  to  the 
vertices  of  the  midsection,  and  to  the  vertices  of  the  pris- 
matoid. The  lines  drawn  from  0  to  the  vertices  of  the 
midsection  divide  the  midsection  into  triangles.  The 
planes  determined  by  0  and  the  edges  of  the  prismatoid 
divide   the   prismatoid   into   two   pyramid's    0-ABC  and 

27 


28 


COMPUTATION   AND    MENSURATION 


0-BEFa,  and  a  series  of  tetrahedrons  0-BDE,  0-BEF, 
0-BFa,  0-BCa,0-OBa,  O-ACB,  and  O-ABB. 

The  volume  of  the  pyramid  0-ABC is  ^  h  -b^  ;  the  vol- 
ume of  the  pyramid  0-BEFGr  is  J  A  •  h^. 


Fig.  9. 


The  triangle  BHK  is  similar  to  the  triangle  BBE 
siud  HK  is  ^  BE.  Hence  area  of  BBE  is  4  times 
area  of  BHK. 

The  tetrahedrons  0-BBE  and  0-BHK  have  the  same 
altitude  and  therefore  are  proportional  to  their  bases. 
Hence  volume  0-BBE=4:  times  volume  0-BHK. 

Tetrahedron  O-^jMT  =  tetrahedron  B-HOK.  The  vol- 
ume of  B-OffK=ih'  OHK.  Hence  volume  0-BBE  = 
^h'OHK. 


VOLUMES    OF    SOLIDS    BOUNDED    BY    PLANES      29 
In  like  manner  it  is  proved  that  the 


volume  of  0-BEF  =  ^h 
volume  of  O-EFG  =  |  A 
volume  of  0-BaC=^h 
volume  of  0-CDa^^h 
volume  of  0-ACD  =^h 
volume  of  O-ABB  =  |  h 


OKL, 

OLM. 

OMK 

ONP. 

OPQ. 

OQH. 


The  sum  of  the  volumes  of  all  these  tetrahedra  is  |^  A  •  m. 

It  follows  that  the  volume  of  the  prismatoid  is 
J  A-(5j  +  ^2  +  4  m),  that  is,  the  volume  of  the  prismatoid  is 
i  of  the  altitude  times  the  sum  of  the  areas  of  the  two 
bases  and  four  times  the  area  of  the  midsection. 

This  result  is  called  the  prismoidal  formula. 

Art.  16.  —  Special  Cases  of  Prismatoids 

When  the  bases  are  equal  polygons,  with  their  corre- 
sponding sides  parallel,  the  prismatoid  is  a  prism. 

When  the  bases  are  similar  polygons,  with  their  corre- 
sponding sides  parallel,  the  prismatoid  is  a  frustum  of  a 
pyramid. 

When  one  base  becomes  a  point,  the  prismatoid  is  a 
pyramid. 

When  one  base  becomes  a  line,  the  prismatoid  is  a 
wedge. 

Wlien  one  face  is  a  rectangle  and  the  bases  are  trape- 
zoids perpendicular  to  the  rectangular  face,  the  prismatoid 
is  called  a  prismoid.  This  is  the  general  shape  of  a  railway 
cutting  or  embankment. 


30  COMPUTATION    AND    MENSURATION 

In  case  of  a  cutting  or  embankment,  when  one  face  is 
not  a  plane,  the  error  in  computing  the  volume  is  dimin- 
ished by  measuring  the  areas  of  2n-\-l  equidistant  cross 
sections  A^^  A^^  A^,  J.^,  ••• ,  A^^^  ^2»+i  ^^^  finding  the  sum 
of  the  volumes  of  the  prismoids  whose  bases  are  A-^  and  J.g, 
A^  and  ^5,  A^  and  ^7,  ••• ,  A^^  and  A^^_^^,  The  approx- 
imation is  made  closer  by  increasing  w. 

Problem  58.  From  the  prismoidal  formula  derive  the 
formulas  for  the  volumes  of  a  pyramid  and  a  frustum  of  a 
pyramid. 

Problem  59.  The  base  of  the  great  pyramid  of  Egypt 
is  a  square  764  feet  on  a  side  and  the  altitude  is  477.6 
feet.     Find  the  volume  in  cubic  yards. 

Problem  60.  The  section  of  a  canal  is  32  feet  wide  at 
the  top,  14  feet  wide  at  the  bottom,  and  8  feet  deep. 
How  many  cubic  yards  were  excavated  in  a  mile  of  the 
canal  ? 

Problem  61.  Find  the  volume  of  a  rectangular  wedge 
whose  base  is  70  meters  by  20  meters,  length  of  edge  110 
meters,  and  altitude  24.8  meters. 

Problem  62.  Find  the  number  of  cubic  yards  of  earth 
excavated  from  a  railway  cutting  made  through  ground 
the  original  surface  of  which  was  an  inclined  plane  run- 
ning in  the  same  direction  as  the  rails.  The  length  of 
the  cutting  is  4  chains  15  links,  the  breadth  at  bottom 
30  feet,  the  breadth  at  top  at  one  end  75  feet,  and  at  the 
other  end  135  feet,  and  the  depths  of  these  ends  20  feet 
and  46  feet  respectively. 


VOLUMES    OF    SOLIDS    BOUNDED    BY   PLANES     31 

Problem  63.  A  railway  embankment  is  to  be  made  on 
ground  which  slopes  20  feet  in  a  mile  in  the  direction  of 
the  rails  and  the  rails  themselves  slope  1  in  700.  The 
embankment  is  straight  for  2^  miles.  The  breadth  at 
the  top  is  29  feet,  the  slopes  of  the  sides  1  in  1,  and  the 
height  at  the  upper  end  is  3  feet.  Find  the  number  of 
cubic  yards  of  earth  required  for  the  embankment. 

Problem  64.  A  railway  cutting  is  to  be  made  30  feet 
wide  at  the  bottom,  the  slopes  of  the  sides  being  1|-  to  1. 
The  depths  in  feet  on  one  side  of  the  cutting,  taken  at 
intervals  of  1  chain,  are  12,  11,  10,  9,  10,  12,  14,  15,  17, 
20,  21,  23,  25,  27,  30,  33,  37,  41,  45,  50,  53  ;  the  corre- 
sponding depths  on  the  opposite  side,  15,  14,  13,  12,  14, 
15,  17,  19,  22,  26,  28,  30,  33,  36,  39,  41,  44,  51,  57,  60,  58. 
Find  the  number  of  cubic  yards  of  earth  to  be  excavated. 


CHAPTER  V 

COMPUTATION  AND  USE  OP  TEIGONOMETEIO  PUNOTIONS 
Art.  17.  —  On  Angles 

The  figure  formed  by  two  straight  lines  proceeding 
from  a  point  is  called  an  angle. 

The  two  straight  lines  are  called  the  sides  of  the  angle, 
and  their  direction  is  indicated  by  arrowheads.  The 
point  is  called  the  vertex  of  the  angle. 

One  side  of  the  angle  is  called  the  initial  side,  the  other 
the  terminal  side  of  the  angle. 


Fig.  10. 

A  circular  arrow  about  the  vertex  between  the  two 
sides  with  the  arrowhead  at  the  terminal  side,  indicates 
which  side  is  to  be  considered  the  terminal  side  of  the 
angle. 

The  angle  is  called  positive  when  this  circular  arrow  is 
anticlockwise  ;  negative  when  the  arrow  is  clockwise. 

When  the  terminal  side  of  the  angle  coincides  with  the 
initial  side  the  angle  is  called  a  perigon.  An  angle  of  one 
degree  is  one  three  hundred  and  sixtieth  of  a  perigon. 

32 


USE    OF    TRIGONOMETRIC    FUNCTIONS  33 

The  ratio  of  the  circuhir  arc,  described  with  the  vertex 
of  the  angle  as  a  center  and  bounded  by  the  sides  of  the 
angle,  to  the  radius  of  the  arc  is  called  the  circular  meas- 
ure of  the  angle. 

Denoting  the  circular  measure  of  the  angle  by  6  and  the 

arc 
radius  of  the  arc  by  r,  it  follows  that  —  =  6  and  arc  =  rd. 


,  Fig.  11. 

When  6  =  1,  arc  =  r  ;  that  is,  the  unit  angle  in  circular 
measure  is  the  angle  at  the  center  of  a  circle  which  inter- 
cepts on  the  circumference  an  arc  equal  in  length  to  the 
radius.  The  unit  angle  in  circular  measure  is  called  the 
radian. 

If  the  angle  is  four  right  angles,  the  arc  is  the  circumfer- 
ence of  the  circle  ;  that  is,  arc  =  2  irr.  Hence,  the  circular 
measure  of  an  angle  of  four  right  angles  is  2  tt  radians. 

It  follows  that  2  TT  radians  are  equivalent  to  360°.    Hence, 

1  radian  = and  1°  =  zr^  radians. 

Problem  65.  Find  the  circular  measure  of  an  angle  of 
6°  correct  to  four  decimal  places. 


34 


COMPUTATION   AND    MENSURATION 


Pboblem  66.  Find  the  circular  measure  of  an  angle  of 
14°  21'  2b"  correct  to  four  decimal  places. 

Problem  67.  Find  in  degrees,  minutes,  and  seconds 
the  angle  whose  circular  measure  is  .357. 

Problem  68.  Find  in  degrees  correct  to  four  decimal 
places  the  angle  1.258. 

Art.  18.  —  Trigonometric  Functions 

Call  the  vertex  of  the  angle  0  and  the  indefinite  straight 
line  in  which  the  initial  side  of  the  angle  lies  XX' ,  Call 
distances  measured  on  XX'  from  0  in  the  direction  of  the 

Y 


Fig.  12. 

arrowhead  on  the  initial  side  positive,  distances  measured 
in  the  opposite  direction  negative.  Let  OX  be  the  pos- 
itive side  of  XX' .  Draw  a  straight  line  YY'  through  0 
perpendicular  to  XX',  If  OF  makes  an  angle  of  +90° 
with  OX,  call  distances  measured  from  0  towards  Y  pos- 
itive and  denote  this  by  the  arrowhead  at  Y, 

Now  take  any  point  P  in  the  terminal  side  of  the  angle. 
The  distance  from  0  to  P  is  called  the  radial  distance  of 
the  point  P  and  is  denoted  by  r.  The  radial  distance  is 
always  positive. 


USE    OF    TRIGONOMETRIC    FUNCTIONS  35 

The  projection  of  the  radial  distance  r  of  the  point  P 
on  the  line  XX'  is  the  abscissa  x  of  the  point  P. 

The  projection  of  the  radial  distance  r  of  the  point  P 
on  the  line  YY'  is  the  ordinate  y  of  the  point  P. 

The  trigonometric  functions  of  an  angle  are  the  follow- 
ing six  ratios : 

The  sine  of  an  angle  is  the  ratio  of  the  ordinate  of  any 
point  in  the  terminal  side  to  the  radial  distance  of  the 

same  point ;  that  is,         sin  0  =  -- 

The  cosine  of  an  angle  is  the  ratio  of  the  abscissa  of 
any  point  in  the  terminal  side  to  the  radial  distance  of  the 

same  point ;  that  is,         cos  0  =  -- 

The  tangent  of  an  angle  is  the  ratio  of  the  ordinate  of 
any  point  in  the  terminal  side  to  the  abscissa  of  the  same 

point ;  that  is,  tan  6  =  -- 

The  cotangent  6  is  the  reciprocal  of  the  tangent  0, 

The  secant  0  is  the  reciprocal  of  the  cosine  0. 

The  cosecant  6  is  the  reciprocal  of  the  sine  6. 

Two  auxiliary  trigonometric  functions  are  the  versed- 
sine  and  coversed-sine,  defined  as  follows  :  versed-sine  6 
=  1  —  cosine  0,  coversed-sine  =  1  —  sine  0. 

Problem  69.  The  tangent  of  an  angle  is  i|.  Con- 
struct the  angle  and  find  the  values  of  all  the  other 
functions  of  the  angle. 

Problem  70  The  sine  of  an  angle  is  .375.  Compute 
the  cosine  and  the  tangent  of  this  angle. 


36  COMPUTATION    AND    MENSURATION 

Problem  71.  The  cosine  of  an  angle  is  —.45.  Com- 
pute the  sine  and  tangent  of  this  angle. 

Art.  19.  —  Computation^  of  Trigonometric  Functions 

If  an  angle  is  given  in  degrees  or  in  radians,  the  angle 
may  be  constructed  by  means  of  a  protractor  and  the 
lengths  of  the  radial  distance,  abscissa,  and  ordinate  of 
any  point  in  the  terminal  side  measured,  and  the  trigo- 
nometric functions  of.  the  angle  computed. 

This  method  does  not  in  general  determine  the  value  of 
the  trigonometric  functions  beyond  the  first  decimal  place. 

If  6  is  the  circular  measure  of  an  angle,  sine  6  may  be 
computed  to  any  required  degree  of  approximation  by 
means  of  the  following  infinite  series,  derived  in  the 
Differential  Calculus, 

^^^  12. 81. 2. 3-4. 5      1.23.4.5.6.7 

+  ...  . 

Let  it  be  required  to  compute  sine  20°  to  four  decimal 
places. 

The  circular  measure  of  20°  is  (9  =  |=  .34907. 

Hence 

6  =  .34907,  f72'73  ^  »QQ707,  1.2.3.4.5  =^ -QQ^Q^. 
It  follows  that  sine  20°  =  .3420. 

The  error  committed  by  omitting  the  terms  of  the  in- 
finite series  after  the  third  may  be  written 

^'*^1.2.3.4.5.6-7L  V8-9     8. 9- 10. 11 7 


USE    OF    TRIGONOMETRIC    FUNCTIONS  37 

'^'  ^^^  1.2.3.t.5.6.7  [-(^  -  8^)-(8.9.lVll 

8 -9. 10- 11. 12.  13;  J 

According  to  (6)  the  error  is  negative,  and  according 

to   (a)   the  numerical   value   of   the    error   is   less   than 

z — 7^ — ^ — ;i — r — ^ — =.     Hence  the  error  committed  by  omit- 

ting  the  terms  of  the  infinite  series  after  the  third  in  com- 
puting sine  20°  is  negative  and  numerically  less  than 
.0000002. 

When  sine  6  has  been  computed,  the  remaining  func- 
tions of  6  may  be  computed  by  means  of  the  fundamental 
relations, 

sin2  ^  +  cos2  (9  =  1,  tan  d  =  ^^^,  cot  6        ^ 


cos  6^  tan  6^ 


sec  0  = -,  cosec  d  — 


cos  d  sin  Q 

The  trigonometric  functions  of  any  angle  wli^tever  may 
be  found  in  terms  of  the  trigonometric  functions  of  an 
angle  not  greater  than  45°.  Therefore  if  the  functions  of 
angles  from  0°  to  45°  are  computed,  the  functions  of  all 
angles  become  known. 

If  Q  denotes  the  circular  measure  of  a  small  angle  and 
Q"  the  number  of  seconds  in  this  angle, 
approximately  sin  0  =  tan  0  =  6  =  6''  x  circular  measure  1" 

20t)264.8* 

The  approximation  is  correct  to  six  places  of  decimals 
from  0°  to  38'  ;  to  five  places  of  decimals  to  1°  20' ;  to 
four  places  of  decimals  to  2°  20". 


38  COMPUTATION    AND    MENSURATION 

Problem  72.  Construct  the  angle  50°  and  find  the 
trigonometric  functions  of  the  angle  by  measurement. 

Problem  73.  Construct  the  angle  110°  and  find  the 
trigonometric  functions  of  the  angle  by  measurement. 

Problem  74.  Compute  the  sine  of  18°  to  four  decimal 
places. 

Problem  75.  Compute  the  sine  of  115°  to  four  deci- 
mal places. 

Problem  76.  Compute  the  sine  of  214°  to  four  deci- 
mal places. 

Problem  77.  Show  that  the  circular  measure,  sine,  and 
tangent  of  any  angle  from  0°  to  7°  are  the  same  to  three 
decimal  places. 

Problem  78.  From  the  general  definitions  of  the 
trigonometric  functions  of  an  angle,  show  that  in  a  right 
triangle,  if  the  hypotenuse  is  c,  the  oblique  angles  A  and 
B  and  the  sides  opposite  a  and  b, 

A       CL  A      b    .        4      a 

smA  =  -,  cos  Az=-,  tan  A  =  ~. 

ceo 

If  the  side  a  is  so  small  compared  with  the  sides  5  and 
c  that  the  angle  A  does  not  exceed  two  or  three  degrees, 


r""^  =  206261:8 '*pp'"°"^'""*^^y' 


A" 
^^  =  20 

and  consequently 

c  =  ^2062^4.8  approximately. 
A. 


USE    OF    TRIGONOMETRIC    FUNCTIONS  39 

Problem  79.  Find  the  area  of  the  equilateral  triangle 
inscribed  in  a  circle  whose  radius  is  15.5  inches. 

Problem  80.  The  angle  at  the  vertex  of  a  cone  is  25°, 
the  altitude  is  12  feet.     Find  the  volume  of  the  cone. 

Problem  81.  A  street  railway  track  is  10  feet  from  the 
curbstone  where  the  track  is  straight.  In  passing  a  corner 
where  the  street  is  deflected  through  an  angle  of  60°,  the 
rail  must  be  4  feet  from  the  corner.  Find  the  radius  of 
the  circular  curve. 

Problem  82.  The  greatest  angle  the  radius  of  the 
earth,  3963.3  miles,  subtends  by  lines  drawn  from  the 
center  of  the  sun  is  8". 8.  Find  the  distance  of  the  sun 
from  the  earth. 

Problem  83.  Two  sides  of  a  triangle  are  76  feet  and 
125  feet,  their  included  angle  is  35°  15'.  Find  the  third 
side  of  the  triangle. 

Problem  84.  The  three  sides  of  a  triangle  are  17.5 
feet,  26.7  feet,  and  35.25  feet.  Find  the  angle  opposite 
the  last  side. 

Problem  85.  Two  sides  of  a  parallelogram  are  125.75 
feet  and  153.25  feet,  their  included  angle  is  67°  45'.  Find 
the  diagonals  of  the  parallelogram. 

Art.  20.  —  On  Vectors 

A  vector  is  a  quantity  which  is  completely  determined 
by  assigning  its  magnitude  and  its  direction. 

A  vector  may  therefore  be  represented  by  a  line  segment 


40 


COMPUTATION    AND    MENSURATION 


and  the  angle  which  this  line  segment  makes  with  a  fixed 
line  of  reference,  the  X-axis,  for  example. 

The  direction  of  a  vector  is  indicated  by  the  arrowhead 
placed  at  the  terminal  point  of  the  line  segment  repre- 
senting the  vector. 

Two  vectors  are  equal  when  they  have  the  same  magni- 
tude and  the  same  direction. 

The  geometric  representative  of  the  vector  will  be  used 

as  the  vector  itself. 

Y 


Fig. 18. 

Let  I  denote  the  length  of  the  vector,  6  the  angle  the 
vector  makes  with  the  X-axis,  x  the  projection  of  the 
vector  on  the  X-axis,  y  the  projection  of  the  vector  on  the 
P-axis.     From  the  figure 

(1)  I  cos  6  =  X,     (2)  Zsin  ^  =  y. 

Squaring  equations  (1)  and  (2)  and  adding, 

l?  =  x'^-\-y\ 

Dividing  (2)  by  (1), 

tan^  =  ^. 


Hence  from  the  projections  of  a  vector  on  the  axes  of 
reference  the  length  and  direction  of  the  vector  may  be 
computed. 


USE    OF    TRIGONOMETRIC    FUNCTIONS 


41 


If  several  vectors  are  drawn  in  such  a  manner  that  the 
initial  point  of  each  successive  vector  is  the  terminal  point 
of  the  preceding  vector,  the  vector  drawn  from  the  initial 
point  of  the  first  vector  to  the  terminal  point  of  the  last 
vector  is  called  the  vector  sum  of  the  several  vectors. 

The  sum  of  the  projections  of  the  several  vectors  on  any 
straight  line  is  the  projection  of  the  vector  sum  on  the 
same  straight  line. 

Let  Zj,  ^2,  ?3,  ?4  denote  the  lengths  of  the  several  vectors, 
dp  ^2'  ^3'  ^4  ^^^^  angles  these  vectors  make  with  the  -X-axis. 


Fig.  14. 


Let  I  denote  the  length  of  the  vector  sum,  0  the  angle 
this  vector  makes  with  the  X-axis,  x  its  projection  on  the 
X-axis,  ^  its  projection  on  the  y-axis. 

From  the  figure 

I  cos  d  =  X=l-^  cos  6^  +  ?2  ^OS  ^2  +  ^3  ^^S  ^3  +  h  ^^^  ^4' 

I  sin  0  =  y  =  1^  sin  6^  +  l^  sin  0^  +  l^  sin  6^  -f- 1^  sin  6^. 

From  these  equations  the  length  and  direction  of  the 
vector  sum  may  be  computed. 

The  displacement  of  a  body  is  determined  by  its  magni- 
tude and  direction.  A  displacement  may  therefore  be 
represented  by  a  vector. 


42  COMPUTATION    AMD    MENSUHATIOK 

If  a  body  has  several  successive  displacements,  each 
displacement  may  be  represented  by  a  vector,  and  the 
resulting  displacement  is  represented  by  the  vector  sum. 

The  force  acting  at  a  point  is  determined  by  its  magni- 
tude and  direction.  A  force  may  therefore  be  represented 
by  a  vector. 

By  experiment  it  is  shown  that  if  several  forces  acting 
at  a  point  are  represented  by  vectors,  the  vector  sum  rep- 
resents the  resultant  of  the  several  forces. 

Problem  86.  Three  sides  of  a  quadrilateral  taken  in 
order  are  11  feet,  1  feet,  and  8  feet,  and  make  angles  of 
18°  18',  74°  50^  and  130°  20'  respectively  with  a  fixed  line. 
Find  the  length  and  direction  of  the  fourth  side  of  the 
quadrilateral. 

Problem  87.  A  man  walks  8, 12, 15,  and  20  miles  along 
successive  straight  lines,  making  angles  respectively  of  30°, 
70°,  120°  15',  and  155°  with  the  E.  W.  line.  Find  the  dis- 
tance and  bearing  of  his  final  position  from  the  starting 
point. 

Problem  88.  The  following  are  the  field  notes  of  a 
surve3^  Bearing  means  the  direction  from  one  station  to 
the  next  succeeding  station. 


Stations 

Bearings 

Distances 

A 

N.  31JW. 

10  chains 

B 

N.  62f  E. 

9.25  chains 

0 

J) 

S.  451  w. 

10.40  chains 

Plot  the  survey,  find  the  distance  and  bearing  from  C 
to  J),  and  find  the  inclosed  area. 


USE  OF  Trigonometric  functions  43 

Problem  89.  Find  the  magnitude  and  direction  of  the 
resultant  of  three  forces  acting  at  a  point  of  11,  7,  and  8 
pounds  and  making  angles  respectively  of  18°  18',  74°  50', 
and  130°  20'  with  a  fixed  line. 

Problem  90.  Find  the  magnitude  and  direction  of  the 
resultant  of  four  forces  acting  at  a  point  of  8,  12,  15,  20 
pounds,  making  angles  respectively  of  30°,  180°,  225°,  330° 
v^^ith  a  fixed  line. 


CHAPTER  VI 

COMPUTATION  AND  USE  OF  LOGAEITHMS 
Art.  21.  —  Natuke  of  Logarithms 

Examine  the  tables, 

I.    100  =  1  11.    IQo    =1 
101  =  10  10-1  =  .1 

102=100  10-2  =.01 

103  =  1000  10-3  =  .001 

104  =  10000  10-4  =  .0001 

105  =  100000  10-5  =  .00001 


Numbers  with  one  integral  place  lie  between  1  and  10, 
and  it  is  evident  that  the  exponent  by  which  10  must  be 
affected  to  give  numbers  between  1  and  10  must  lie  be- 
tween 0  and  1. 

In  like  manner  it  becomes  evident  that  the  exponent 
by  which  10  must  be  affected  to  give  a  number  with  two 
integral  places  must  lie  between  1  and  2  ;  to  give  a  num- 
ber with  three  integral  places  the  exponent  must  lie 
between  2  and  3  ;  and  in  general  to  give  a  number  with 
n  integral  places  the  exponent  must  lie  between  n  —  1 
and  n. 

Decimal  fractions  with  the  first  significant  figure  in  the 
first  decimal  place  lie  between  1  and  .1,  and  it  is  evident 

44 


COMPUTATION   AND    USE    OF   LOGARITHMS        45 

from  table  II  that  the  exponent  by  which  10   must   be 
affected  to  give  numbers  between  1  and  .1  must  lie  be-* 
tween  0  and  —  1. 

In  like  manner  it  becomes  evident  that  the  exponent  by 
which  10  must  be  affected  to  give  a  decimal  fraction  with 
the  first  significant  figure  in  the  second  decimal  place 
must  lie  between  —  1  and  —  2  ;  to  give  a  decimal  fraction 
with  the  first  significant  figure  in  the  third  decimal  place, 
the  exponent  must  lie  between  —  2  and  —  3  ;  and  in  gen- 
eral to  give  a  decimal  fraction  with  the  first  significant 
figure  in  the  ^th  decimal  place,  the  exponent  must  lie 
between  —(n—  1)  and  —  n. 

The  exponent  by  which  10  must  be  affected  to  give  any 
number  is  called  the  logarithm  of  that  number  to  base  10. 

The  logarithm  to  base  10  of  a  number  is  called  the 
common  logarithm  of  that  number  and  is  denoted  by 
writing  log^^  before  the  number.  For  example,  log^Q  100 
=  2,logi„.01=-2. 

Problem  91.  If  the  base  of  a  system  of  logarithms  is 
3,  what  are  the  logarithms  of  27,  243,  2V?  ^^-^? 

Problem  92.    Can  1  be  used  as  the  base  of  a  system  of 

logarithms  ? 

Problem  93.  Can  —  5  be  used  as  the  base  of  a  system 
of  logarithms  ? 

Art.  22.  —  Computation  of  Common  Logarithms 

(a)  Let  ?^J,  n^,  ^3,  n^  represent  the  logarithms  of  the 
numbers  iVj,  JV^,  iVg,  iV^  to  base  10. 

By  definition  JV^  =  10"s  iV^  =  10«2,  iVg  =  10«3,  JST^  =  lO"*. 


46  COMPUTATION    AND    MENSURATION 

The  product  of  the  numbers  iV^,  JV^,  iVg,  iV^,  by  the  law 
of  indices  in  Algebra,  is 

Hence  by  the  definition  of  the  logarithm, 
log  (iV^iiVa^glV^)  =  ^^  +  ^2  +  ^3  +  n^ 

=  log  iV\  +  log  JSr^  +  log  JV3  +  log  iV^, 

that  is,  the  logarithm  of  the  product  equals  the  sum  of  the 
logarithms  of  the  factors. 

(5)  Any  number  in  the  decimal  notation  may  be  writtea 
as  a  common  fraction  whose  numerator  is  integral  and 
whose  denominator  is  some  power  of  10.     For  example, 

375.485  =  ^llll^  =  375485  x  10-3. 

Applying  the  rule  just  proved  to  this  product, 
log  875.485  =  log  375485  +  log  10-3. 

As  a  consequence  of  (5)  the  direct  computatiqn  of 
the  logarithms  of  whole  numbers  only  is  necessary. 

As  a  consequence  of  (a)  the  direct  computation  of  the 
logarithms  of  prime  numbers  only  is  necessary. 

The  logarithms  to  base  10  of  the  prime  numbers  may 
be  computed  to  any  required  degree  of  approximation  by 
using  the  infinite  series 
(1)     logio(l  +  ^)  =  logio^  +  2(0.43429448)^^-ip^  + 

1 + L_+ 1 +  ...+ 

3(2a;+ 1)3^5(22^  +  1)5      7(2a;+l)7 

1  1 


(2  7i-l)(2  2J  +  l)2«-i      (2n  +  l)(2  a; +1)2^+1 
which  is  derived  in  the  Differential  Calculus. 


+  •■ 


COMPUTATION    AND    USE    OF   LOGARITHMS 


47 


The  application  of  this  formula  requires  the  logi^a:,  that 
is,  the  logarithm  of  the  number  one  less  than  the  given 
prime,  to  be  known. 

Let  it  be  required  to  compute  the  common  logarithms 
of  2,  4,  5  to  five  decimal  places. 

1.    Place  a:  =  1  in  (1).     There  results 


log,„  2  =  log,,  1  +  2(0.43429448)g  +  _!_  +  _!_ 

+     1      • 


7-3' 


1 


The  computations  are  carried  out  to  six  decimal  places 
in  order  to  obtain  the  correction  in  the  fifth  decimal 
place.  The  following  is  a  convenient  arrangement  of  the 
computation.  Begin  by  computing  the  sum  of  the  series 
in  brackets. 


^=0.333333 

i  =  0.388333 
0 

1  =  00.87037 
8^ 

1.1  =  0.012246 

1  =  0.004115 
35 

1.1  =  0.000823 
5    3^ 

1  =  0.000457 

11  =  0.000065 

1  =  0.000051 
39 

11=0.000006 

P  =  0.000006 

^^.1_  =  0.000001 

1  =  0.000007 

0.346574 

48  COMPUTATION   AND    MENSUBATION 

The  sum  of  all  the  omitted  terms  of  the  bracketed  in- 
finite series  is 

13    313^15     315^17     317^19    3i9 


which  is  less  than 


L3    313V         32^34     36^      J 


13 

The   series    in    parenthesis    is    an    infinite    decreasing 
geometric   progression,  whose  sum  is  found   by  the    for- 
mula S=  -^-  to  be  -. 
1  —  r  8 

Hence  the  sum  of  all  the  omitted  terms  of  the  original 
infinite  series  is  less  than 

^•p- ^  =  0.00000006, 

and  does  not  influence  the  sixth  decimal  place. 
It  follows  that 

logio2  =  logio  1  +  2  (0.43429448)(0.346574) 
=  0.30103. 

2.  Iogio4  =  21ogio  2  =  0.60206. 

3.  logio  5  =  logio  (1  +  4)  =  logio  4  +  2  (0.43429448) 

ri+i.i+i.i+i.i+i.i+...i 

[9      3     93^5     9^     7    97^9     99         J 

Computation  of  sum   of  infinite  series  to  six  decimal 
places: 


COMPUTATION    AND    USE    OF   LOGARITHMS        49 


?:  =0.111111 

9 

-  =  0.111111 

1  =  0.001372 
93 

1 
3 

.  13  =  0.000457 

1  =  0.000017 
95 

1 
5 

.1  =  0.000003 
9^ 

0.111571 

1  =  0.0000002 
9' 

The  error  committed  by  omitting  all  the  terms  after  the 
third  is 

1    1.1.  1  +  1. 1+1. J_+... 

7'  97^9    99      11     911     13    913^      ' 

which  is  less  than 

1 .  Vl  +  1  +  1  + 1  +  . . .^  =  0.00000C03, 

7  97V    92   94  ^  9«     y 

and  does  not  affect  the  sixth  decimal  place. 
It  follows  that 

logio5  =  logio4  -h  2(0.43429448)(0.111571) 
=  0.60206  +  0.9691 
=  0.69897. 

Problem  94.  Compute  log^^  3. 

Problem  95.  Compute  log^^  7. 

Problem  96.  Compute  logio  11. 

Problem  97.  Compute  logjo  13. 

Problem  98.  Compute  logio  17. 


50  COMPUTATION    AND    MENSURATION 

Art.  23.  —  Arrangement  of  Tables  of  Logarithms 

The  common  logarithm  of  a  number  consists  of  two 
parts,  an  integer  and  a  decimal  fraction. 

From  a  table  of  five-place  logarithms  it  is  found  that 
logio  97756  =  4.99014,  hence  lO^-saoi*  =  97756. 

Dividing  the  last  equation  by  10  several  times  in  suc- 
cession, 

104.99014  ^  97750  lience  log^^  97756  =  4.99014 

103.99014  _  9775.6  hence  log^^  9775.6  =  3.99014 

102.99014  _  977.56  hence  log^^  977.56  =  2.99014 

101.99014  _  97.756  hence  log^^  97.756  =  1.99014 

100.99014  ^  9.7756  hence  log^^  9.7756  =  0.99014 

101.99014  ^  0.97756  hence  log^^  0.97756  =  1.99014 

102.99014  ^  0.097756  hence  logj^  0.097756  =  2.99014 

103.99014  ^  0.0097756  hence  logj^  0.0097756  =  8.99014 

104.99014  ^  0.00097756      hence  log^^  0.00097756  =  4.99014 

The  minus  sign  placed  over  the  1,  2,  3,  4  in  the  last 
four  logarithms  denotes  that  only  the  integral  part  of  the 
logarithm  is  negative.  The  decimal  part  of  the  logarithm 
is  always  positive. 

This  manner  of  writing  the  logarithm  of  a  number  is 
adopted  in  order  that  the  decimal  part  of  the  logarithm 
shall  depend  only  on  the  significant  figures  of  the  number 
and  be  independent  of  the  position  of  the  decimal  point  in 
the  number. 

When  the  logarithm  is  w^ritten  in  this  manner,  the  in- 
tegral part  is  called  the  characteristic;  the  decimal  part, 
the  mantissa. 

The  characteristic  is  determined  by  the  rule: 


COMPUTATION   AND    USE    OF   LOGARITHMS        51 

If  the  number  is  greater  than  1,  the  characteristic  is 
positive  and  less  by  unity  than  the  number  of  integral 
places  in  the  number. 

If  the  number  is  less  than  1,  the  characteristic  is  neg- 
ative and  equal  to  the  number  of  the  decimal  place 
occupied  by  the  first  significant  figure  of  the  number. 

The  mantissa  of  the  logarithm  of  a  number  is  taken 
from  the  table  of  logarithms. 

If  five-place  logarithms  are  used,  it  is  absurd  to  attempt 
to  compute  beyond  six  significant  figures,  and  even  the 
sixth  figure  and  frequently  the  fifth  figure  is  unreliable. 
The  truth  of  this  statement  will  become  evident  from  the 
computations  themselves. 

Examplp:  1.  Let  it  be  required  to  find  the  common 
logarithm  of  375.658  from  the  five-place  table. 

By  rule  the  characteristic  is  2.  The  mantissa  corre- 
sponding to  the  first  four  figures  of  the  number  is  taken 
directly  from  the  table.  The  logarithms  of  the  two  con- 
secutive numbers  of  four  significant  figures  between  which 
the  given  number  lies  are 

log  375.7  =  2.57484 
log  375.6  =  2.57473 

The  difference  in  the  mantissa  corresponding  to  1  unit 
of  the  fourth  significant  figure  of  the  number  is  11  units 
of  the  fifth  decimal  place  of  the  mantissa. 

Assuming  that  the  mantissa  increases  uniformly  while 
the  number  increases  by  one  unit  of  the  fourth  significant 
figure,  it  follows  that  the  increase  of  the  mantissa  due  to 
the  fifth  figure  of  the  number  is  .5  x  11  =  5.5  units  of  the 


62  COMPUTATION    AND    MENSUBATION 

fifth  decimal  place,  and  the  increase  due  to  the  sixth  figure 
of  the  number  is  .08  x  11  =  .88  units  of  the  fifth  decimal 
place  of  the  mantissa. 
Collecting  the  results, 

'^^^^f'^^F^N^  log  375.6     =2.57473 

^'VERSITY  J  increase  for  .05    =    .000055 

^^UFORhU^V^  increase  for  . 008  =    .0000088 
""^  Ince  log  375.658  =  2.57479 

Observe  that  if  any  figure  other  than  zero  stood  in  the 
sixth  significant  place  of  the  number,  the  logarithm  of  the 
number  to  five  decimal  places  would  remain  the  same. 

The  method  of  finding  the  corrections  of  the  logarithm 
for  the  fifth  and  sixth  figures  of  the  number  is  called 
interpolation. 

Example  2.  Let  it  be  required  to  find  the  number 
whose  common  logarithm  is  3.72564. 

The  characteristic  shows  that  there  are  four  integral 
places  in  the  number  required.  The  figures  of  the  number 
corresponding  to  the  mantissa  are  found  from  the  table  of 
logarithms. 

Take  from  the  table  the  logarithm  next  lower  than  the 
given  logarithm  and  subtract  it  from  the  given  logarithm 
and  from  the  next  higher  logarithm. 

3.72564  3.72567  =  log  5307 

3.72558  =  log  5306  3.72558  =  log  5306 

6  9 

A  difference  of  9  units  in  the  fifth  decimal  place  of  the 
mantissa  gives  a  difference  of  1  unit  in  the  fourth  signifi- 


.  COMPUTATION    AND    USE    OF   LOGARITHMS         53 

cant  figure  of  the  number.  Assuming  that  the  rate  of 
change  of  the  mantissa  from  log  5306  to  log  5307  is  uni- 
form, the  increase  in  the  number  due  to  the  increase  of 
the  mantissa  by  6  units  in  the  fifth  decimal  place  is 
6-i-9  =  .67  units  of  the  fourth  significant  figure  of  the 
number.     Hence  3. 72564  =  log^,  5306. 67. 

The  sixth  figure  of  this  number  is  uncertain.  It  may 
have  any  value  from  2  to  9.  This  makes  the  uncertainty 
in  the  number  extend  to  the  fifth  significant  figure,  which 
may  be  either  6  or  7  if  the  number  is  to  be  expressed 
correct  to  five  significant  figures. 

It  appears  from  these  examples  that  the  use  of  five-place 
logarithmic  tables  limits  the  computation  to  four  or  five 
significant  figures. 

Problem  99.  From  a  five-place  table  find  the  logarithm 
of  1476.38.  What  other  numbers  of  six  figures  have  the 
same  logarithm  ? 

Problem  lOO.  From  a  five-place  table  find  the  logarithm 
of  8754.88.  What  other  numbers  of  six  figures  have  the 
same  logarithm  ? 

Problem  loi.  From  a  five-place  table  find  the  number 
whose  logarithm  is  2.14896  and  determine  the  limits  of 
uncertainty  oi  the  number. 

Problem  102.  From  a  five-place  table  find  the  number 
whose  logarithm  is  4.79658  and  determine  the  limits  of 
uncertainty  of  the  number. 

Problem  io3.  How  does  the  change  in  the  mantissa 
corresponding  to  a  change  of  1  unit  in  the  fourth  signifi- 
cant figure  in  the  number  vary  with  the  number  ? 


54  COMPUTATION    AND    MENSURATION 

Art.  24. — Computation  by  Means  of  Logarithms 

Logarithms  are  used  to  shorten  the  operations  of  multi- 
plication, division,  involution,  and  evolution. 

These  operations  are  performed  by  applying  the  follow- 
ing rules : 

Let  logjQ  M=  m,  logj^  iV=  n.     By  definition 

M  =  w\  jsr=  io\ 

The  product  of  ilff  aiid  iVis 

Hence  by  definition 

logio  (MJST)  =m-{-n  =  logj^  M-^  ^ogio^;  that  is, 

I.  The  logarithm  of  the  product  equals  the  sum  of  the 
logarithms  of  the  factors. 

The  quotient  of  il[f  and  iVis 


jsr 

Hence  by  definition 
M 


—  =  10"^~". 


logio  7i?  =  ^^^  ~  ^  =  ^^^10  ^-  ^^Sio  ^>  that  is, 


II.    The  logarithm  of  the  quotient  equals  the  logarithm 
of  the  dividend  minus  the  logarithm  of  the  divisor. 

Raising   both   sides   of   the   equation   M=  IC"   to  the 
power  jt?. 

Hence  by  definition 

logio  MP  =  p w  =  p  logiQ  M  5  that  is, 


COMPUTATION    AND    USE    OF   LOGARITHMS        55 

III.  The  logarithm  of  the  power  of  a  number  equals 
the  logarithm  of  the  number  multiplied  by  the  exponent 
of  the  power. 

Extracting  the  r  root  of  both  sides  of  the  equation 


1  m 


Hence  by  definition 


log,, if L  !?^  =  1  logio ^;  that  is, 
r      T 

IV.  The  logarithm  of  the  root  of  a  number  equals 
the  logarithm  of  the  number  divided  by  the  index  of 
the  root. 

In  working  a  numerical  problem,  first,  decide  on  a 
method  of  computation ;  second,  arrange  a  scheme  of  com- 
putation;  third,  perform  the  computation. 

Example.  Let  it  be  required  to  compute  the  lateral 
area  and  the  volume  of  a  cone  of  revolution,  altitude  29.75 
feet,  radius  of  base  12.15  feet. 

Denoting  the  lateral  area  by  A^  the  volume  by  FJ  the 
radius  of  the  base  by  i2,  and  the  altitude  by  IT, 


(a)  A  =  'TrB  ^E^-\-E% 
(J)  V=  1  ttR^H. 
Applying  logarithms  to  («), 

log  J.  =  log  TT  +  log  R  +  1  log  (^  +  E^). 


56  COMPUTATION   AND    MENSURATION 

Scheme  of  computation  : 

log  ^  =  1.47349  i?2^885.08 

log  H^  =  2. 94698  i^  =  147. 62 

log  ^2  =2.16916  E^  +  ff^  =  10S2.10 

log7r  =  0.49715 

+  log  72  =  1.08458 

+  1  log  (^2  +  ji2^  =  1.50698 

log  v4  =3.08871 

^  =  1226. 63  square  feet. 

Applying  logarithms  to  (5), 

log  r=  log  TT  +  2  log  R  +  log  H-  log  3. 

Scheme  of  computation : 

log  7r  =  0.49715 

2  log  i2=  2.16916 

log  ff=  1.47349 

4.13980 

-  log  3  =  0.47712 

log  F=  3. 66268 

r=  4599.2  cubic  feet. 

In  these  values  of  A  and  Fthe  fifth  and  sixth  figures 
are  uncertain. 

Problem  i04.  The  chord  of  a  circular  segment  of  one 
base  is  3.25  inches,  the  altitude  of  the  segment  is  1.15 
inches.     Find  the  diameter  of  the  circle. 

Problem  io5.  Find  the  area  of  the  surface  of  the 
sphere  whose  radius  is  6.7  inches. 


.    COMPUTATION    AND    USE    OF   LOGARITHMS         57 

Problem  io6.  Find  the  volume  of  the  sphere  whose 
diameter  is  15.36  feet. 

Problem  107.  Find  the  area  of  the  equilateral  triangle 
whose  side  is  27.16  feet. 

Problem  1O8.  Find  the  capacity  in  gallons  of  a  rec- 
tangular cistern  whose  dimensions  are  10.5,  6.35,  and 
4:.idb  feet. 

Problem  io9.  Find  the  capacity  in  gallons  of  a  tank 
in  the  shape  of  a  circular  cylinder,  radius  of  base  8.35  feet, 
altitude  12.46  feet. 

Problem  110.  Find  the  volume  of  a  regular  pyramid 
whose  base  is  a  square  4.18  feet  on  a  side  and  whose 
lateral  edge  is  11.27  feet. 

Problem  111.  The  radius  of  a  circle  is  8.57  feet.  Find 
the  least  chord  that  can  be  drawn  through  a  point  3.25 
feet  from  the  center. 

Problem  112.  The  diameter  of  a  circle  is  15.28  inches. 
Find  the  length  of  the  tangent  to  the  circle  from  a  point 
20.15  inches  from  the  center. 

Problem  113.  In  a  right  triangle  the  sides  about  the 
right  angle  are  5.75  feet  and  8.17  feet.  Find  the  perpen- 
dicular from  the  vertex  of  the  right  angle  to  the  hypot- 
enuse. 

Problem  114.  An  excavation  1.5  yards  deep,  rectan- 
gular at  top  and  bottom,  and  in  the  form  of  a  frustum  of 
a  pyramid  has  its  upper  base  10  yards  wide  and  16  yards 
long,  and  the  lower  base  7.5  yards  wide.  How  many  cubic 
yards  were  removed  to  make  the  excavation  ? 


58  COMPUTATION    AND    MENSUBATION 

Problem  115.  If  the  atmosphere  extends  to  a  height 
of  45  miles  above  the  earth's  surface,  what  is  the  ratio 
of  its  volume  to  the  volume  of  the  earth,  assuming  the 
earth  to  be  a  sphere  with  a  diameter  of  7912  miles. 

Problem  116.  The  circumference  of  the  base  of  a  conic 
frustum  is  49.3  feet,  the  diameter  of  the  top  is  12.5  feet, 
the  altitude  is  15.7  feet.     Find  the  surface  and  volume. 

Problem  117.  On  a  sphere  whose  radius  is  11.75 
inches,  what  is  the  area  of  a  zone  whose  altitude  is  3.25 
inches  ? 

Problem  118.  On  a  sphere  whose  radius  is  28.5  feet 
find  the  area  of  a  spherical  triangle  whose  angles  are  120°, 
130°,  and  140°. 

Problem  119.  On  a  sphere  whose  diameter  is  25  inches 
find  the  area  of  the  spherical  polygon  whose  angles  are 
110°,  120°,  130°,  and  140°. 

Problem  120.  The  sides  of  a  triangle  are  each  12.49 
inches.  What  is  the  volume  of  the  solid  generated  by 
revolving  the  triangle  about  one  side  ? 

Problem  121.  A  dome  is  in  the  form  of  a  spherical 
zone  of  one  base.  Its  height  is  30  feet  and  the  diameter 
of  the  base  60  feet.     Find  its  area. 

Problem  122.  Find  the  volume  of  the  frustum  of  the 
cone  of  revolution,  the  radii  of  whose  bases  are  12.5  feet 
and  7.25  feet  and  whose  slant  height  is  9.75  feet. 

Problem  123.  The  base  of  a  triangle  is  Q.b  feet,  the 
altitude  is  15  feet.     Find  at  what  distances  from  the  base 


COMPUTATION    AND    USE    OF   LOGARITHMS         59 

lines  parallel  to  the  base  must  be  drawn  to  divide  the  tri- 
angle into  three  equal  areas. 

Problem  124.  The  radius  of  the  base  of  a  cone  is  7.2 
feet,  the  altitude  is  25  feet.  Find  at  what  distances  from 
the  base  planes  parallel  to  the  base  must  be  drawn  to 
divide  the  cone  into  three  equal  volumes. 

Problem  125.  Two  parallel  planes  divide  the  diameter 
of  a  sphere  into  three  equal  parts.  The  radius  of  the 
sphere  is  12  feet.  Find  the  ratio  of  the  volumes  of  the 
solids  into  which  the  sphere  is  divided. 

Art.  25.  —  The  Compound  Interest  Formula 

If  a  sum  of  money  P  is  invested  at  compound  interest 
at  R  per  cent,  the  amount  at  the  end  of  the  first  year  is 
P(l  +  R). 

This  amount  is  the  principal  at  the  beginning  of  the 
second  year,  hence  the  amount  at  the  end  of  the  second 
year  is  P(l  +  R)\ 

In  like  manner  the  amount  at  the  end  of  the  third  year 
is  found  to  be  P(l  +  Ry. 

In  general  the  amount  at  the  end  of  t  years  is 

^  =  P(l  +  Ry. 

From  this  it  follows  that  the  present  worth  of  a  sum  of 
money  A  due  t  years  hence,  if  money  is  worth  R  per  cent 
compound  interest  is 

P  =  A(1  +  RyK 

Example  1.  What  is  the  present  worth  of  an  annual 
pension  of  $500  to  run  for  7  years,  the  first  payment  to. 


60  COMPUTATION    AND    MENSUBATION 

be  made  one  year  from  date,  and  money  being  worth  5  % 
compound  interest  ? 

The  present  worth  of  the  first  payment  is  500  x  1.05"^ 
The  present  worth  of  the  second  payment  is  500  x  1.05-2 
The  present  worth  of  the  third  payment  is  500  x  1.05-^ 
The  present  worth  of  the  fourth  payment  is  500  x  1.05"* 
The  present  worth  of  the  fifth  payment  is  500  x  1.05"^ 
The  present  worth  of  the  sixth  payment  is  500  x  1.05"^ 
The  present  worth  of  the  seventh  payment  is  500  x  1.05"^ 
The  present  worth  of  the  seven  payments  is 

F  =  500[1.05-i  +  1.05-2  +  1.05-3  +  1.05-*  +  1.05-^ 

+  1.05-6  + 1.05-7] 

rnnl.05-8-  1.05-1        rnAl.05-7  -  1 


=  ^^^     1.05- 

1-1 

•^""  1  -  1.05 

^,,,1-105-7 
.05 

=  10000(1  - 

1.05-7). 

To  compute  x 

=  1.05-7, 

,  pass  to  logarithms, 

loga;  = 

-7  log  1.05 

= 

-0.14833 

= 

1.85167. 

.'.X  = 

0.710667. 

It  follows  that 

P=  10000(1  -0.710667) 
=  12893.33. 

Example  2.  In  how  many  years  will  an  annual  pay- 
ment of  $300  meet  principal  and  interest  of  a  debt  of 
f  2000,  money  being  worth  5  %  compound  interest  ? 


COMPUTATION   AND    USE    OF  LOGARITHMS        61 

Let  t  represent  the  required  number  of  years.  At  the 
end  of  t  years  the  debt  amounts  to  2000  x  1.05^ 

If  the  first  annual  payment  is  to  be  made  at  the  end  of 
the  first  year,  its  amount  at  the  end  of  t  years  is  300 
X  1.05'~i.  The  amount  at  the  end  of  t  years  of  the 
second  annual  payment  is  300  x  1.05'~2,  and  so  on  for  the 
successive  payments.  The  last  annual  payment  is  to  be 
made  at  the  end  of  the  t  years  and  its  value  then  is  #300. 

Hence  the  sum  of  the  amounts  of  the  t  annual  payments 

is 

A  =  300[1  + 1.05  4- 1.052  +  1.053  +  ...  +  1.05'-i] 

=  30oli^^^^  =  6000(1. 05^ -1). 

At  the  end  of  t  years  the  amount  of  the  debt  is  to  equal 
the  amount  of  the  annual  payments.     Hence 

2000  X  1.05'  =  6000(1.05'  -  1). 

Solving  this  equation  for  1.05', 

1.05' =  1.5. 

Passing  to  logarithms, 

^log  1.05  =  log  1.5. 

Solving  for  t, 

^^logL5      3  3^ 
log  1.05  ^ 

Problem  126.  What  is  the  present  worth  of  82000 
due  10  years  hence  without  interest,  if  money  is  worth 
6  %  compound  interest  ? 

Problem  127.  A  man  pays  a  premium  of  8104  per 
annum  on  a  life  policy  of  $3500  for  twenty  years  before 


62  COMPUTATION    AND    MENSURATION 

death.     Money  being  worth  5  %  compound  interest,  does 
the  insurance  company  gain  or  lose,  and  how  much  ? 

Problem  128.  A  town,  whose  property  has  the  as- 
sessed valuation  of  $7,325,000  and  whose  annual  tax  rate 
is  19  mills  on  fl,  issues  bonds  to  the  amount  of  $225,000 
to  build  a  sewerage  system,  these  bonds  being  at  4|  per  cent 
interest  and  maturing  in  15  years.  An  extra  tax  is  to  be 
laid  to  meet  this  interest  and  to  provide  a  sinking  fund  to 
redeem  the  bonds,  the  rate  of  interest  in  the  sinking  fund 
being  3 J  per  cent  compounded  annually.  How  many 
mills  must  be  added  to  the  tax  rate  for  this  purpose  ? 

Problem  129.  A  town  finds  it  necessary  to  build  a 
bridge.  The  cost  of  a  stone  bridge  is  $20,000,  the  cost  of 
maintenance  150  per  year,  the  life  of  the  bridge  50  years. 
The  cost  of  a  steel  bridge  is  $10,000,  the  cost  of  main- 
tenance $150  per  year,  the  life  of  the  bridge  15  years. 
The  town  can  borrow  money  at  4  per  cent  compound  in- 
terest and  realize  4  per  cent  compound  interest  on  its 
sinking  fund.  What  is  the  cost  per  year  to  the  town  of 
each  of  the  two  bridges  ? 

Art.  26.  —  The  Slide  Eule 

If  line  segments  are  laid  off  from  one  point  of  a  straight 
line  proportional  to  the  mantissas  of  the  logarithms  of 
numbers  from  100  to  1000,  and  the  number  whose  loga- 
rithm is  represented  by  the  line  segment  is  written  over 
the  terminal  point  of  the  line  segment,  there  will  be  formed 
a  logarithmic  scale. 

P'rom  a  logarithmic  scale  nine  inches  long  constructed 


COMPUTATION    AND    USE    OF   LOGARITHMS         63 

in  this  manner  the  line  segment  which  represents  the  loga- 
rithm of  a  number  of  three  figures,  and  conversely  the 
number  to  three  figures  whose  logarithm  is  represented  by 
a  given  line  segment,  can  be  read  off  with  considerable 
accurac}^ 

To  find  the  product  of  two  numbers,  apply  to  the  loga- 
rithmic scale  the  sum  of  the  line  segments  representing 
the  logarithms  of  the  numbers.  Over  the  terminal  point 
of  the  sum  of  these  line  segments  stands  the  number  which 
is  the  product  of  the  two  given  numbers. 


I  I  I  I  1 1 1 1  itiiMiiiiiliiiihiiiliiiiiiiiiMiiiiiiiiiUM^ 


Fig.  15. 

To  find  the  quotient  of  two  numbers,  apply  to  the  loga- 
rithmic scale  the  difference  of  the  line  segments  repre- 
senting respectively  the  dividend  and  divisor.  Over  the 
terminal  point  of  the  difference  of  these  line  segments 
stands  the  number  which  is  the  quotient  of  the  two  given 
numbers. 

The  slide  rule  consists  of  two  logarithmic  scales,  one  of 
which  slides  along  the  other. 

To  find  the  product  of  two  numbers  by  means  of  the 
slide  rule,  locate  one  number  on  the  fixed  logarithmic 
scale  and  bring  the  initial  point  of  the  sliding  scale  over 
this  number.  Now  locate  the  second  number  on  the  slid- 
ing scale  and  in  line  with  this  number  on  the  fixed  scale 
the  product  is  found. 

To  find  the  quotient  of  two  numbers  by  means  of  the 
slide  rule,  locate  the  dividend  on  the  fixed  scale  and  the 


64  COMPUTATION    AND    MENSURATION 

divisor  on  the  sliding  scale  and  bring  these  two  num- 
bers over  each  other.  The  quotient  is  the  number  on  the 
fixed  scale  in  line  with  the  initial  point  of  the  sliding 
scale. 

To  find  the  square  root  of  a  number  by  means  of  the 
slide  rule,  bisect  the  line  segment  which  represents  the 
logarithm  of  the  number  and  find  the  number  whose  loga- 
rithm is  represented  by  the  corresponding  line  segment. 

In  like  manner  any  root  of  a  number  may  be  found  by 
means  of  the  slide  rule. 


ar^  —  a         a 

ar^ 

r-1        1-r 

1-r' 

a               ar^ 

CHAPTER  YII 

ON  LIMITS 

Art.  27.  —  The  Infinite  Decreasing  Geometric 
Progression 

The  formula  for  the  sura  of  the  first  n  terms  of  a  geo- 
metric progression  whose  first  term  is  a  and  ratio  r  is 

whence 

'-      1-r  1-r 

In  these  formulas  a  and  r  for  a  particular  progression 
have  fixed  values;  that  is,  they  are  constants  ;  while  n  may 
be  any  positive  integer  and  the  value  of  aS'^  depends  on  n; 
that  is,  n  and  S^  are  variables. 

If  r  is  less  than  unity  and  n  increases  without  limit, 
the  progression  becomes  an  infinite  decreasing  geometric 
progression. 

The  difference  between  the  variable  /^„  and  the  constant 

— - —  is  numerically  equal  to .     When  r  is  less  than 

1-r  -^    ^  1-r 

one,  if  any  numerical  value,  however  small,  be  assigned  in 
advance,  a  value  of  n  can  be  computed  such  that  for  this 
p  66 


66  COMPUTATION    AND    MENSURATION 

value  of  n  and  all  larger  values  of  n  the  difference  between 

the  variable  S^  and  the  constant  :; shall  be  less  than 

1  —  r 

the  value  assigned. 

This   is   expressed   by   saying   that    the   limit   of    the 
variable  S^  when  n  is  infinitely  increased  is  the  constant 

,  and  it  is  written 


1-r 


limit  aSJ  =     ^ 


1-r 

In  general,  if  a  variable  approaches  a  constant  in  such  a 
manner  that  the  numerical  value  of  the  difference  between 
the  variable  and  the  constant  becomes  and  remains  less 
than  any  value,  however  small,  that  can  be  assigned  in 
advance,  the  variable  is  said  to  approach  the  constant  as 
its  limit. 

Pkoblem  130.  Find  the  limit  of  the  sum  of  the  first  n 
terms  of  1  +  |-  +  -|  +  2V  +  FT  +  *"  when  n  is  infinitely 
increased. 

Problem  131.  For  what  values  of  n  does  the  sum  of 
the  first  n  terms  of  1  +  J  +  ^  +2V  +  "sS:  +  *'*  ^i^^^  from 
the  limit  of  the  sum  by  less  than  .00001  ? 

Akt.  28.  —  The  Length  of  a  Curved  Line 

The  approximate  length  of  a  curved  line  is  found  by 
inscribing  or  circumscribing  a  broken  straight  line  the 
parts  of  which  fit  the  corresponding  parts  of  the  curved 
line  as  closely  as  may  be  practicable.  The  approximation 
is   made   closer  by  diminishing  the  lengths  of  the  parts 


ON    LIMITS  67 

of  the  broken  line.  The  limits  of  accuracy  of  direct 
measurement  are  soon  reached. 

The  length  of  a  curved  line  is  defined  as  the  limit  of  the 
inscribed  or  circumscribed  broken  line  when  the  number 
of  parts  of  the  broken  line  is  infinitely  increased  in  such  a 
manner  that  the  length  of  each  part  of  the  broken  line  is 
infinitely  diminished. 

By  using  this  definition  the  length  of  a  curved  line  may 
be  computed  either  exactly  or  to  any  required  degree  of 
approximation. 

Art.  29.  —  The  Computation  of  tt 
By  definition 

circumference  of  circle      semicircumference 


TT 


diameter  radius 


Hence  if  the  radius  is  made  unity,  tt  equals  the  length  of 
the  semicircumference. 

If  (?„  represents  the  length  of  the  side  of  the  regular 
polygon  of  n  sides  inscribed  in  the  circle,  and  f„  represents 
the  side  of  the  regular  polygon  of  n  sides  circumscribed 
about  the  circle,  the  semicircumference  always  lies  between 
^nCn  and  ^nt„  and  the  semicircumference  is  the  common 
limit  of  ^  ncJ^  and  |  nt^  when  n  is  infinitely  increased. 

It  is  readily  proved  that 

(1)  t„  =  — % 


(2)  c^n  =  V2  -  -V-i-cl 

The  side  of  the  inscribed  regular  hexagon  is  the  radius 
of  the  circle.     Hence  Cq=  1.     By  formula  (1)  the  side 


68  COMPUTATIOIf    AND    MENSUBATION 

of  the  circumscribed  regular  hexagon,  ^g,  is  computed. 
By  formula  (2)  the  side  of  the  inscribed  regular  polygon 
of  12  sides,  e^^^  is  computed. 

By  repeated  successive  application  of  formulas  (1)  and 
(2)  the  sides  of  inscribed  and  circumscribed  regular 
polygons,  the  number  of  whose  sides  is  six  multiplied  by 
any  power  of  two,  may  be  computed. 

The  results  of  twelve  successive  applications  of  for- 
mulas (1)  and  (2)  give  the  following  table  of  values  for 
the  semiperimeters  of  the  polygons; 


n 

\nc,. 

Intn 

6 

3.0000000 

3.4641016 

12 

3.1058285 

3.2153903 

24 

3.1326286 

3.1596599 

48 

3.1393502 

3.1460862 

96 

3.1410319 

3.1427146 

192 

3.1414524 

3.1418730 

384   " 

3.1415576 

3.1416627 

768 

3.1415836 

3.1416101 

1536 

3.1415904 

3.1415970 

3072 

3.1415921 

3.1415937 

6144 

3.1415925 

3.1415929 

12288 

3.1415926 

3.1415927 

The  semicircumference  of  the  circle,  and  therefore  tt, 
always  lies  between  the  semiperimeters  of  the  inscribed 
and  circumscribed  regular  polygons  of  the  same  number 
of  sides. 

Hence  the  value  of  tt  correct  to  six  decimal  places  is 
3.141593.  . 


ON    LIMITS  69 

By  direct  measurement  the  value  of  tt  could  not  be 
determined  beyond  the  second  or  third  decimal  place. 
There  is  no  limit  to  the  approximation  that  may  be 
reached  by  computation. 

Art.  30. — An  Important  Limit 

The  derivation  of  many  of  the  formulas  for  the  deter- 
mination, exact  or  approximate,  of  lengths,  areas,  and 
volumes  is  based  on  the  limit,  when  n  is  infinitely  in- 
creased, of  the  ratio  of  the  sum  of  the  p  powers  of  the  first 
n  natural  numbers  to  the  p  +  1  power  of  w,  or  expressed 
as  a  formula, 

IP  +  2^  4-  3^  +  4^  +  5^  -h  •••  +  nP 


limit 


tP+i 


If  p  is  any  positive  integer,  and  if  a  >  6, 

^ "^ —  =  aP  +aP-^b  +  aP-^^  +  •••  +  aJ^-i  +  h^ 

a  —  0 

which  may  be  written 

(1)  CP  +  i)b''<^^^^^^. 

a  —  0 
From  (1),  by  placing  h  successively  equal  to 
1,  2,  3,  4,  5,  ••• ,  w  and  a  =  J  +  1, 
(p  +  1)  1^  <  2^+1  -  1^+1, 
(;?+"l)  2^  <  3^+1  -2^+1, 
(p  4-  1)  3^  <  4^+1  -  3^+1, 
{p  4-  1)  4^  <  5^+1  -  4^+1, 


(j9  4- 1)  ^^  <  (w  +  1)^+1  -  ri^+i. 


70  COMPUTATION   AND    MENSURATION 

Adding  these  n  inequalities, 
(^p  +  1)(1^  +  2^  +  3^  +  4^  +  •••  +  w^)  <  (n+  1)^+1  -  1^+^ 
Dividing  the  last  inequality  by  (jt?  +  l)n^+i, 
IP  +  2^  +  3^  +  4^+  •••  -f  nP         1      r  A      1  Yl^_l_ I 

<     1     {l+^J  +  £C|±_l)+...+iL_±l}, 
jt?  +  1  I  n  2!7i2  n^-i   J 

and  finally 

Ip  4-  2^  4-  3^  +  4^  4-  •••  -^n^  1 

^P+i  JO  4-  1 


< 


jt?  +  1  I     w  2!?i' 


n^ 


p-i 


If   now  9i  is   infinitely   increased,  since  -  becomes   in- 

n 

finitely  small  and  the  number  of  terms  in  the  bracket  is 
jt?  —  1,  a  finite  number, 

limit  1^  +  2^  +  3^  +  4^4-  •••  +71^ ^ _1_ 

i?+  1* 


n=oo  ^^+1 


Art.  31. — Lexgth  and  Area  of  Involute  of  Circle 

If  a  fine  inextensible  string,  coiled  around  a  circle,  be 
uncoiled  from  the  circle  and  the  string  be  always  kept 
taut,  the  free  extremity  of  the  string  will  trace  the  curve 
called  the  involute  of  the  circle. 

Denote  the  radius  of  the  circle  by  r  and  suppose  the 
length  of  the  string  uncoiled  to  be  rd.  Divide  the  arc  r6 
into  n  equal  parts  and  inscribe  in  the  arc  a  regular  broken 


ON    LIMITS  71 


line.     Denote  by  a  the  length  of  each  part  of  the  broken 

line.     Then  a  =  2r  sin  -— .      If   the  string   were   wound 

2n 

around  the  broken  line  and  then  uncoiled,  if  S^  represents 


Fig.  16. 

the  length  of  the  curve  traced  by  the  extremity  of  the 
string  and  A^^  the  area  swept  over  by  the  string, 

A  n  \ji       n        n        n  n  J 

^„=2  7'2sin2--    -H 1 +  ...  H . 

zn\ji       n         n  7i  j 

When  n  is  infinitely  increased,  the  broken  line  approaches 
the  circular  arc  rd  as  its  limit,  and  the  involute  of  the 
broken  line  approaches  the  involute  of  the  circular  arc  as 

its  limit. 

6  6 

Since  limit  sin—-  =  limit  -— ,  it  follows  that 
74=00         'zn       n=»    2n 

limit  S.  =  re^  limit  p +  2  +  3  +  4  +  .■■  +>»1,g,^ 

limit  4„ = 1  ,-^^3  ii„,it  ri^+2^+3^+f+-+>»n  ^  i  ,.^. 

n=«  n=oo     L  ^  J 


72  COMPUTATION   AND    MEN8UBATI0N 

Problem  132.    Find  the  length  and  area  of  the  involute 
of  the  circle  radius  r  extending  from  ^  =  0  to  ^  =  2  tt. 

Problem  133.    Find  the  length  and  area  of  the  involute 
of  the  circle  radius  r  extending  from  ^  =  27rto  ^  =  4  7r. 


CHAPTER  VIII 

GEAPHIO  ALGEBEA 
Art.  32. — The  Graph  of  an  Equation- 

If  an  equation  /(a;,  y)  =  0  is  satisfied  by  the  coordinates 
of  every  point  (x^  y)  of  a  line,  and  if  every  point  (a:,  ?/) 
whose  coordinates  satisfy  the  equation  is  a  point  of  the 
line,  the  line  is  called  the  graph  of  the  equation. 

To  construct  the  graph  of  an  equation /(ic,  y)  =  0,  com- 
pute the  values  of  y  for  different  values  of  a;,  and  locate 
the  points  whose  coordinates  are  the  pairs  of  corresponding 
real  values  of  x  and  y.  The  graph  of  the  equation  is  the 
smooth  curve  drawn  through  the  points  located. 

Example.    Construct  the  graph  of  x^-\-y^  —  ^. 

The  corresponding  pairs  of  real  values  of  x  and  y  are 

a;=_3     -2  -1  0     +1  +2  +3 

y=   0  ±2.24  ±2.83  ±3  ±2.83  ±2.24    0. 

For  a;  >  +  3  and  for  x<  —  3,  ?/  is  imaginary.  Hence  the 
graph  lies  between  the  lines  a:  =  3  and  a:  =  —  3.  The 
graph  also  lies  between  the  lines  ^  =  3  and  «/  =  —  3. 

Locating  the  points  whose  coordinates  are  the  corre- 
sponding pairs  of  values  of  x  and  y  and  drawing  a  smooth 
line  through  these  points,  it  becomes  evident  that  the 
graph  resembles  a  circle. 

73 


74 


COMPUTATION    AND    MENSURATION 


In  fact  the  form  of  the  equation  shows  at  once  that  the 
graph  is  a  circle  whose  radius  is  3  and  center  the  origin. 


\ 

r 

^ 

■^ 

/ 

\ 

1 

0 

\ 

^  X 

\ 

\ 

J 

v 

y 

Notice  that  for 
the  circle  x^  +  y"^- 
the  circle  aP'-\-y'^- 
without  the  circle 

Problem  134. 

Problem  135. 

Problem  ise. 


Fig.  17. 

all  points  (x^  y)  on  the  circumference  of 
-9  =  0;  that  for  all  points  (a;,  y)  within 
-  9  <  0  ;  and  that  for  all  points  (a:,  ^) 
a:2  +  /-9>0. 

Construct  the  graph  of  3a^— 4«/  =  12. 

Construct  the  graph  of  x^-\-y'^  —  16. 

Construct  the  graph  of  —  +  ^  =  1. 


Problem  137.    Construct  the  graph  of 


!-■ 


Problem  138.    Construct  the  graph  of  ?/2  =  4  x. 
Problem  139.    Construct  the  graph  of  :c2  =  4  y. 


GBAPHIC    ALGEBRA  lb 

Problem  140.    Construct  the  graph  oi  y  =  Zx—  ba^. 
Problem  141.    Construct  the  graph  oiy  =  b  —  x-\-Qx^, 

Art.  33.  —  Equations  of  Lines 

Suppose  the  relation  between  the   coordinates  of  any 
point  P  (x^  y)  to  be  expressed  by  the  first  degree  equation 

(1)  Ax^  By  -^  C=^. 

Let  PiCa^i,  «/i),  PgC^  ^2)^  A(%  2/3)  be  any  three  points 
whose  coordinates  satisfy  this  equation,  that  is, 

Ax^  j^By^-{-  0=0, 
Ax^  +  %2  +0=0, 
Ax^  +  %3  +  C  =  0. 

The  elimination  of  A,  B^  C  gives 

^1(^2  -  ^3)  +  ^2(^3  -  ^i)  +  ^3(^1  -  ^2)  =  ^• 
The  left-hand  member  of  this  equation  is  double  the 
area  of  the  triangle  whose  vertices  are  P^,  Pg'  ^3-  ^^ 
follows  that  the  area  of  the  triangle  whose  vertices  are 
any  three  points  whose  coordinates  satisfy  equation  (1)  is 
zero;  that  is,  these  three  points  lie  in  a  straight  line. 
Hence  the  graph  of  every  first  degree  equation  is  a 
straight  line. 

The  equation 

(2)  0.-2  +  ^2  =  ^2 

expresses  the  fact  that  the  distance  from  the  origin  to  the 
point  (x,  y)  is  a.  Therefore  all  points  (x^  y\  whose  co- 
ordinates satisfy  equation  (2)  are  located  on  the  circle 
whose  radius  is  a  and  center  at  origin. 


76  COMPUTATION   AND    MENSURATION 

From  equation  (2) 

y  =  Va2  —  ic^. 

Multiplying  y  by  _,  that  is,  multiplying  each  ordinate  of 

the  circle  by  the  same  factor  -, 

a 


a 
which  reduces  to 

The  curve  formed  by  the  points  (x^  y^  whose  coordi- 
nates satisfy  equation  (3)  is  called  the  ellipse. 

The  curve  formed  by  the  points  (.t,  y)  whose  coordinates 
satisfy  the  equation 

(4)  ^  ^-^'  =  1 

is  called  the  hyperbola. 

The  curve  formed  by  the  points  {x^  y)  whose  coordi- 
nates satisfy  the  equation 

(5)  y'^  =  1'px 
is  called  the  parabola. 

Problem  142.  Show  that  the  area  of  the  ellipse 
^  +  f-  =  1  is  iTob. 

Art.  34.  —  Graphic  Solution  op  Equations 

Example  1.  Solve  the  equation  a^  —1  x  +  1  =  0 
graphically. 

Construct  the  graph  oi  y  =  x^  —  1  x  -{-1, 


QttAPHtC   ALGEhBA 


n 


The  abscissas  of  the  points  where  the  graph  intersects 
the  JT-axis  are  the  roots  of  the  equation  x^  —  1  x  +  l  =  0. 
These  roots  are  found  by  measurement  to  be  1.3, 1.7,  — 3.1. 


Y 

t          1 

__:n:__± 

^..., 

-\-\- 

::;:::  ::|:: 

-\-v- 

:-::o:;:i: 

mm 

■^x 


> 

< 

:;z 

/\ 

^^ 

^^  '   ■ 

7 

V   ^^ 

7 

7       ^ 

r 

f       4 

0 

X- 

r       / 

\ 

A         ' 

ST 

\/ 

^z 

\ 
\ 

±. 

Fig.  18.  Fig.  19. 

Example  2.  Solve  graphically  the  simultaneous  equa- 
tions ^2  =  10  a;  and  x^  +  ^'■^  =  25. 

Construct  the  graphs  of  these  two  equations  on  the 
same  axes  of  reference.  The  coordinates  of  the  points  of 
intersection  of  the  graphs  are  the  common  solutions  of  the 
equations.  The  points  of  intersection  are  (2.1,  4.4)  and 
(2.1,-4.4). 

Problem  143.    Solve  graphically  x^  —  4:x  —  15  =  0, 

Pkoblem  144.    Solve  graphically  a^-\-7x— 7  =  0. 

Problem  145.    Solve  graphically  o^  -^  y^  z=  25, 

^2=  10  a;  —  a^. 

Problem  146.    Solve  graphically  rr^  +  ?/2  =  25, 

z^-if  =  4. 


78 


COMPUTATION    AND    MENSURATION 


Art.  35.  —  Inequalities  treated  Graphically 

Example.  Show  graphically  for  what  pairs  of  values 
of  X  and  y  the  inequalities 

X  -  y  -\-l>0 
2x-\-  y  -Q>0 
-^x-\-2y-\-  10>0 
are  true  ? 

Construct  the  graphs  of  the  equations 

The  graph  of  the  equation  x  —  y  ■\- 1  ==  0  divides  the 
XF'-plane  into  two  parts,  such  that  for  the  coordinates  of 
every  point  (x^  ?/)  in  one  part  x  —  y  +  1>  0  and  for  the 


Fig.  20. 


coordinates  of  every  point  (x^  y)  in  the  other  part  x  —  y 
+  1  <0.  Determine  by  trial  for  which  part  x  —  y  -\-l>0 
and  indicate  this  part  of  the  plane  by  the  arrowhead 
placed  on  the  graph. 


GRAPHIC    ALGEBRA  79 

Kepeat  the  same  operation  with  the  other  two  graphs. 
The  shaded  triangle  in  the  figure  contains  all  the  points 
(^x,  y)  whose  coordinates  satisfy  simultaneously  the  three 
given  inequalities. 

Problem  147.  Solve  graphically  the  simultaneous  in- 
equalities 2— ?/  +  4>0,  a:— 2?/  +  5<0,  2;  +  5y  —  4<0. 

Problem  148.  Solve  graphically  the  simultaneous  in- 
equalities 2;2  -f  ^2  _  25  <  0,  2  2;  +  y  -  10  >  0. 


CHAPTER   IX 

AEEAS  BOUNDED  BY  OUEVES 
Art.  36. — Exact  Areas 

If  the  equation  of  a  curve  is  known,  the  ordinate  corre- 
sponding to  any  abscissa  may  be  computed. 

If  a  curve  is  given  whose  equation  is  not  known,  the 
ordinate  corresponding  to  any  abscissa  may  be  measured. 

Denote  by  a  and  h  the  abscissas  of  the  end  points  of  a 
curve  and  assume  h>a.  Divide  h  —  a  into  m  equal  parts 
and  call  each  part  A,  so  that  mh  =  b  —  a. 

Denote  by  ^q,  y^,  i/^,  y^,  •  •  •  ,  ?/^  the  ordinates  of  the  points 
of  the  curve  whose  distances  from  a  are  the  multiples  of  h. 


■►X 


Fig.  21. 


Construct  rectangles  on  each  successive  ordinate  and 
the  adjacent  part  oil  —  a  located  to  the  right  of  the  re- 
spective ordinate. 

80 


AREAS    BOUNDED    BY    CURVES  81 

The  limit  of  the  sum  of  the  areas  of  these  rectangles 
when  m  is  infinitely  increased  is  the  area  bounded  by  the 
curve,  the  ordinates  of  the  end  points  of  the  curve  and 
the  X-axis. 

If' the  relation  between  ordinate  and  abscissa  is  ex- 
pressed by  an  equation  of  the  form 

(1)  7/  =  A-i-Bx-\-C2^-hI>:i^-{-  •••  -\-]Srx\ 

this  limit  can  be  determined  exactly. 

Consider  the  rectangle  constructed  on  the  ordinate  2/1 
corresponding  to  the  point  whose  distance  from  a  is  Ih. 
The  area  of  this  rectangle  is 

Substituting  for  A  its  value     ~    ,  the    expression   for 

m 

the  area  bounded  by  the  curve  (1),  the  X-axis,  and  the 

ordinates  corresponding  to  x=^a^  x=h  becomes 


l=m-\ 


Area  =  limit  V 


A(h-a)--\-B(h-ay 


I 


m 


and  finally 

Area  =  ^  (/,  _  a)  +  |  (6  -  a^  +  ^(h  -  a^ 


N    ^7        Nw+i 


+  ...+      (j_«) 


+ 

When  w  =  1,  equation  (1)  represents  a  straight  line  and 
the  area  is  given  by  the  formula 

Area  =  A(h  -  a)  +  f  (6  -  a)2. 


82  COMPUTATION    AND    MENSURATION 

When  71  =  2,  equation  (1)  represents  a  parabola  and  the 
area  is  given  by 

Area  =  A(h-a)  +  ~(h  +  ay-{-  ^(h  +  a^. 

Problem  149.  Find  the  area  bounded  by  ^  =  3-f-52;, 
a;  =  2,  rr  =  10,  and  ?/  =  0. 

Pkoblem  150.  Find  the  area  bounded  by  ?/  =  2  2:  +  6  a:^^ 
a;  =  0,  a?  =  5,  and  «/  =  0. 

Problem  151.  Find  the  area  bounded  by  x^  =  2py, 
a;  =  0,  x=  a^  and  ?/  =  0. 

Art.  37. — Approximate  Equation  of  a  Curve 

If  the  equation  of  a  curve  is  not  of  the  form  (1),  it  may 
be  possible  to  express  approximately  the  relation  between 
ordinate  and  abscissa  by  means  of  an  equation  of  the  form 
(1)  and  thus  obtain  an  approximate  value  of  the  area. 

For  example,  the  equation  of  the  circle  is 

x^  +  y'^  =  r% 
from  which 

,     ^/^ 2        n      1'-^^      1    Srr*      1    3    5a;6  -] 

^  L2r^24r4246r6  J 

This  infinite  series  is  convergent  for  all  values  of  x 
numerically  less  than  r,  hence  for  all  such  values  y  can  be 
expressed  in  the  form  A+  Bx-\-  Cx^  + 1>^  +  •  •  •  +  iVic"  to 
any  required  degree  of  approximation,  and  consequently 
the  area  of  the  circular  segment  may  be  computed  to  any 
desired  degree  of  approximation  by  the  methods  of  the 
previous  article. 

If  a  curve  is  given  whose  equation  is  not  known,  and  if 
the  abscissas  of  the  end  points  are  x  —  a  and  x  =  h^  divide 


AREAS    BOUNDED    BY    CURVES  83 

h  —  a  into  n  equal  parts  and  call  each  part  k^  so  that 
h  —  a  =  nk.  The  ordinates  corresponding  to  the  points  of 
division  may  be  measured.  Denote  by  y^,  y^,  y^.  Vz^---  ^Vn 
the  ordinates  corresponding  to  the  abscissas  a,  a  +  ^,  a  +  2  ^, 
a  +  3  A:,  •  •  • ,  0^  +  w^. 

It  is  possible  to  determine  an  equation  of  the  form 

(1)  y^^A^Bx^Cx^^D:^^-  ■■•  +  Nx\ 

which  is  satisfied  by  the  coordinates  of  these  n-{-l  points 
of  the  curve. 

For  the  successive  substitution  of  the  coordinates  of  the 
n-\-l  points 

(a,  yo),  (a4-/c,  y{),  (a -{-2k,  ^/g),  •••,  (a  +  wA:,  ?/„) 

in  equation  (1)  furnislies  n-\-l  equations  of  the  first 
degree  between  the  n  -{-  1  coefficients  of  equation  (1). 
These  equations  determine  the  coefficients. 

In  general  the  higher  the  degree  of  equation  (1),  and 
consequently  the  greater  the  number  of  points  the  line 
represented  by  equation  (1)  has  in  common  with  the 
given  curve,  the  more  closely  will  the  line  represented  by 
equation  (1)  fit  the  given  curve  and  the  closer  will  be  the 
approximation  of  the  area  bounded  by  equation  (1)  to  the 
area  bounded  by  the  given  curve. 


Art.  38.  —  Approximate  Areas 
First  approximation.    When   w  =  1,    equation    (1)    be- 
comes 

y  =  A  +  Bx, 

and  Area  =  A(h  —  a)  +  —(b  —  a^. 


84  COMPUTATION    AND    MENSURATION 

Denoting  the  ordinates  of  the  end  points  of  the  given 
curve  by  ^^  and  ^^^ 

yi  =  ^^ 

y^  =  A  +  B(h-  a). 

From  these  equations  A  '=  y-^^  B  =  ~     7         • 
Substituting  these  values, 

Area  =  -^  (^^  +  ^2)- 

This  is  the  approximation  found  by  replacing  the  given 
curve  by  a  straight  line  joining  the  end  points  of  the 
curve. 

Second  approximation.  When  ?^  =  2,  equation  (1)  be- 
comes 

y  =  A^Bx-{-  Qx\ 

^^^  Area  =  vl(5  -  «)  +  f  (5  -  aj'  +  ^(h  -  a^. 

Denoting  the  three  equidistant  ordinates  of  the  given 
curve  by  y^,  y^,  y^, 

y,  =  A^^Ch-a}-\-^Cb-ay, 

y^  =  A  +  B(h-a)+  Q(h  -  ay. 
From  these  equations 
^  =  Vv 

h  —  a 
(5  -  af         • 


AREAS    BOUNDED    BY    CURVES  85 

Substituting  these  values, 

Area  =  —^  (t/^  +  4  ^2  +  ^s)- 

This  is  the  approximation  found  by  replacing  the  given 
curve  by  a  parabola  through  three  points  of  the  curve 
with  equidistant  ordinates.  It  is  known  as  Simpson's 
approximation  for  the  area  bounded  by  a  curve. 

Third  approximation.  When  n  =  S,  equation  (1)  be- 
comes 

and 

Area  =  AQh  -  a) +~(h  -  ay+^^b  -  ay  +  ~(h  -ay, 

Z  O  4: 

Denoting  the  four  equidistant  ordinates  of  the  given 
curve  by  i/^,  y^,  y^,  and  y^, 

y^  =  A^  \B(h  -  «)  +  \C(h  -  ay  +  i^BQy  -  ay, 
y^  =  A  +  lB(h  -  a)  +  I  C(h  -  ay  +  ^^j  D(h  -  ay, 
y^  =  A  +  B(ib-a)-^  C(h  -  ay  +  D(h  -  ay. 

Solving  these  equations  for  A,  B,  (7,  D  and  substituting, 

Area  =\(h-a)  {y^  +  8(^3  +  y^)  +  y^. 

Other  approximations.  When  ri  =  4  or  5,  the  resulting 
expression  for  the  approximate  area  does  not  take  a 
simple  form. 

When  w  =  6,  if  the  equidistant  ordinates  are  denoted 
^y  Vv  Vv  ^3'  ^4^  Vh'  y%>  Vv  ^^^  expression  for  the  area  is 
Area  =  -^^(h  +  a)  {y^  +  ^^  +  ^^  +  y^  +  5(^2  +  i/^)  +  6  y^  }. 

This  is  Weddle's  approximation  for  the  area  bounded 
by  a  curve. 


86  COMPUTATION    AND    MENSURATION 

The  application  of  these  several  formulas  for  areas  will 
not  give  close  approximations  if  the  equidistant  ordinates 
have  a  wide  range  of  values.  It  is  therefore  important  to 
break  up  the  given  curve  into  parts  such  that  for  each 
part  the  equidistant  ordinates  have  a  small  range  of  values 
and  apply  the  formulas  to  each  part  separately. 

Problem  152.  By  using  the  value  of  tt  compute  correct  to 
two  decimal  places  the  area  of  the  segment  of  the  circle  whose 
radius  is  14  feet  bounded  by  a  diameter  and  a  chord  parallel 
to  the  diameter  and  distant  7  feet  from  the  diameter. 

Problem  153.  Compute  six  equidistant  ordinates  of 
the  circular  segment  of  Problem  149  and  find  the  error 
of  the  first  approximation. 

Problem  154.  Using  the  six  equidistant  ordinates  of 
Problem  150,  find  the  error  of  Simpson's  approximation. 

Problem  155.  Compute  seven  equidistant  ordinates  of 
the  circular  segment  of  Problem  149  and  find  the  error 
of  Weddle's  approximation. 

Problem  156.  Find,  approximately,  the  area  of  a  plot 
of  ground  bounded  on  one  side  by  a  straight  line  and  on 
the  other  by  an  irregular  curve.  The  length  of  the  straight 
line  is  8.97  chains,  and  the  lengths  in  chains  of  31  equi- 
distant ordinates  taken  in  order,  measured  from  this  line 
to  the  opposite  boundary,  are 

1.37  1.83  2.31  '   2.52 

2.46  2.73  2.91  3.21 

1.98  1.74  1.50  1.59 

2.93  3.41  3.92  3.42 

1.75  1.50  1.65 


2.58 

2.47 

2.32 

2.75 

2.63 

2.01 

1.98 

2.37 

2.76 

2.76 

2.42 

2.01 

CHAPTER  X 

VOLUMES   OF  SOLIDS 
Art.  39.  —  Exact  Volumes 

Suppose  two  parallel  planes  to  be  drawn  such  that  the 
solid  extends  from  the  one  plane  to  the  other  and  is  con- 
tained wholly  between  the  two  planes. 

Divide  the  distance  between  these  two  planes  into  n 
equal  parts,  and  through  the  points  of  division  pass  planes 
parallel  to  the  planes  drawn  at  the  start. 

Starting  from  one  of  the  original  parallel  planes,  con- 


FiG.  22. 


struct  cylinders  on  the  successive  plane  sections  of   the 
solid  as  bases  with  one  of  the  n  equal  parts  of  the  distance 
between  the  original  parallel  plahes  as  altitude. 
Denoting  by  Xq,  X^,  X^,  Xg,  X^,  • 

87 


X,i  the  areas  of 


88  COMPUTATION    AND    MENSURATION 

the  successive  equidistant  sections  of  the  solid,  and  the 
distance  between  the  parallel  planes  including  the  solid 
hj  h  —  a^  the  sum  of  the  volumes  of  the  cylinders  is 

It 

and  the  volume  of  the  given  solid  is  the  limit  of  this  sum 
when  n  is  infinitely  increased. 

If  the  areas  of  the  successive  plane  sections  of  the  solid 
can  be  expressed  in  terms  of  the  distance  x  from  one  of 
the  end  planes  by  an  equation  of  the  form 

X  =  A  +  Bx-\-Cx^  +  I)3?-\-Ex^+  ••.  +iV^% 
the  limit  of  the  sum  can  be  determined  exactly.     In  fact 
the  determination  of  this  limit  is  identical  with  the  deter- 
mination of  the  limit  for  areas  of  Article  36. 

Art.  40.  —  Approximate  Volumes 
The  volume  of  a  solid  is  given  by  the  expression 
Volume  = 

iimit(Xo+Xi+  X2+X3+X4+  -  +x„_i)^i:^. 

«  =  Q0  71 

It  follows  that  the  approximation  to  the  volume  is  made 
closer  by  increasing  the  value  of  n. 

First  approximation.  When  n=l,  denoting  by  X^  and 
X2  the  areas  of  the  end  sections, 

Volume  =  ^  (Xj  +  X^). 

Second  approximation.  When  ^  =  2,  denoting  by  Xj,  X^^ 
and  X3  the  areas  of  the  three  equidistant  sections, 

Volume  =  ^  (Xi  +  4  X^  +  X3). 


VOLUMES    OF   SOLIDS  89 

This  is  the  prismoidal  formula. 

Third  approximation.  When  w  =  3,  denoting  the  areas 
of  the  four  equidistant  sections  by  X^,  X^,  X^,  and  X^, 

Volume  =  1  (5  -  a)  [X^  +  ^CX^  +  X^)  +  XJ. 

Fourth  approximation.  When  w  =  6,  denoting  the  areas 
of  the  seven  equidistant  sections  by  Xj,  Xg,  Xg,  X^,  Xg, 

Xg,  Xy, 

Volume  = 

Problem!  157.  The  solid  generated  by  revolving  a  cir- 
cular segment  of  one  base  about  its  chord  is  called  a 
circular  spindle.  Find  the  first,  second,  third,  and  fourth 
approximations  of  the  volume  of  the  circular  spindle  if  the 
radius  of  the  circle  is  14  feet  and  the  height  of  the  circular 
segment  is  T  feet. 

Problem  158.  A  vessel  laden  with  a  cargo  floats  at 
rest  in  still  water  and  the  line  of  flotation  is  marked. 
Upon  the  removal  of  the  cargo  every  part  of  the  vessel 
rises  3  feet,  when  the  line  of  flotation  is  again  marked. 
From  the  known  lines  of  the  vessel  the  areas  of  the  two 
planes  of  flotation  and  of  five  equidistant  sections  are 
calculated  and  found  to  be  as  follows,  expressed  in  square 
feet: 

3918        3794        3661        3517       3361        3191        3004 

Find  the  weight  of  the  cargo  removed,  supposing  355 
cubic  feet  of  water  to  weigh  10  tons. 


90  COMPUTATION    AND    MENSURATION 

Art.  41.  —  Applicability  of  Prismoidal  Formula 
The  prismoidal  formula 

Volume  =  l(h-  a)(Xi  +  4 Xg  +  X3) 
gives  the  exact  volume  of  any  solid  bounded  by  two  par- 
allel planes  provided 

X=A-\-Bx^  Cx\ 
where  X  denotes  the  area  of  a  section  of  the  solid  made 
by  a  plane  parallel  to  the  end  planes,  x  denotes  the  dis- 
tance from  this  plane  section  to  one  of  the  end  planes,  and     ^ 
A,  B^  Q  are  constants. 

a.    For  prisms  and  cylinders 
X=A 
and  the  prismoidal  formula  applies. 
h.    For  pyramids  and  cones 

X=Cx^ 
and  the  prismoidal  formula  applies. 

c.  For  frustums  of  pyramids  and  cones 

X=^A  +  Bx  +  Cx^ 
and  the  prismoidal  formula  applies. 

d.  For  a  sphere 


X=  iri^ax-x^) 


Fig.  23. 

and  the  prismoidal  formula  applies. 


VOLUMES    OF    SOLIDS  91 

e.    For  an  ellipsoid  of  revolution  or  spheroid 

or 
and    the    prismoidal    formula 
applies. 

Fig.  24. 


^.    For  any  ellipsoid 
Y 


X=7r-^C2ax-x^) 

and     tlie     prismoidal 
formula  applies. 


Fig.  25. 


g.    For  the  paraboloid  of  revolution 
Y 


X=  2pirx 
>-X    and    the    prismoidal    for- 
mula applies. 


Fig.  26. 


92  COMPUTATION    AND    MENSURATION 

Problem  159.  The  radius  of  a  sphere  is  10  feet.  Find 
the  volume  of  the  spherical  segment  whose  bases  are  dis- 
tant 2  feet  and  4  feet  from  the  center  of  the  sphere. 

Problem  leo.    Find  the  volume  of  the  solid  generated 

2       2 
by  revolving  the  ellipse  ~-\-^=l  about  the  rc-axis. 

Problem  I6I.  Find  the  volume  of  the  ellipsoid  whose 
axes  are  10,  14,  18  inches. 

Problem  162.  Find  the  volume  of  the  solid  generated 
by  revolving  about  the  a;-axis  that  part  of  the  parabola 
y2  =  2jt?a;  bounded  by  a:  =  0  and  x  =  a. 

Problem  163.  Find  the  volume  of  the  sphere  whose 
radius  is  R, 

Problem  164.  Find  the  volume  of  the  frustum  of  a 
cone  of  revolution  whose  altitude  is  A,  the  radii  of  the 
bases  M  and  r. 

Problem  165.  A  cask  is  constructed  in  the  shape  of 
the  middle  frustum  of  a  spheroid.  The  inside  dimensions 
of  the  cask  are  length  1^  meters,  end  diameters  |  meter, 
bung  diameter  ^  meter.  Find  the  capacity  of  the  cask  in 
liters  and  in  gallons. 


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